Introduction Ideals, Varieties, and Algorithms Lecture 1homepages.warwick.ac.uk/staff/D.Maclagan/AARMS/AARMSTropical… · What is tropical geometry? First answer: (Algebraic) geometry

AARMS TROPICAL GEOMETRY

DIANE MACLAGAN

1. Introduction

These notes are my lecture notes from a four week graduate summer school onTropical Geometry held at the University of New Brunswick in July/August 2008under the auspices of the Atlantic Association for Research in the MathematicalSciences (AARMS). The course was aimed at graduate students finishing their firstyear of graduate studies, though in practice a quarter of the class was about to startgraduate school, and a few were more experienced. The only official prerequisitesare a solid course in abstract algebra, and linear algebra. Recommended backgroundreading was Ideals, Varieties, and Algorithms [CLO07], by Cox, Little, and O’Shea.

These are lecture notes, so are not attempting to be complete, both in contentand in references. In particular, these notes only cover one aspect of this excitingemerging field - search for “tropical geometry” in mathscinet or on the arXiv to seemuch much more.

There are almost certainly errors, both typographical, and mathematical (bothminor and major) in these notes. Please let me know (D.Maclagan at warwick.ac.uk)any mistakes you find. Thanks are due to AARMS 2008 students for the typosalready corrected!

2. Lecture 1

What is tropical geometry?First answer: (Algebraic) geometry where instead of working over the complexnumbers (or some other field) we work over the tropical semiring (R,⊕,⊗). Here ⊕is the usual minimum, and ⊗ is the usual addition.Example: (5⊕ 6)⊗ 7 = 12.

When necessary we consider (R ∪ ∞,⊕,⊗), so ∞ becomes the additive identity.Then (R∪∞,⊕,⊗) is a semiring (ie associative, distributative etc - just no additiveinverse).

In algebraic geometry we often work with polynomials. In tropical geometry we“tropicalize” these polynomials, which turns them into piecewise linear functions.Example: f(x, y) = x2 + y2 − 1. This tropicalizes to trop(f) = x2 ⊕ y2 ⊕ 0 =min(2x, 2y, 0). This is a piecewise linear function. (See below for why the −1 turnsinto 0). See Figure 1.

In algebraic geometry we study the common zeros of polynomial equations (Warn-ing: Oversimplification!). These are called varieties. Tropically this corresponds totaking the nonlinear locus of the polynomial trop(f).Example: Let f(x, y) = x + y + 1. The variety of f is the set {(x, y) ∈ C2 :x+y+1 = 0}, which is a line in C2. Then trop(f) = min(x, y, 0). This is a piecewise

1

2 DIANE MACLAGAN

0

2y

2x

Figure 1.

linear function with graph shown in Figure 2. The nonlinear locus is the three linesegments x = y ≤ 0, x = 0 ≤ y, and y = 0 ≤ x. This is also shown in Figure 2.

y = 0 ≤ x

0

y

x

x = 0 ≤ y

x = y ≤ 0

Figure 2.

Thus varieties turn into polyhedral complexes under the tropicalization map.Warning: An issue with this first answer is that not everything tropicalizes well. Inparticular, maps between varieties do not tropicalize precisely as expected. (“Trop-icalization is not functorial”). For this reason we will be careful over the next twoweeks to define things formally.Motivation

Why tropicalize? A first reason is that polyhedral geometry is (often) easier thanalgebraic geometry. Many invariants of the variety become invariants of the resultingpolyhedral complex.Example: Let f = x + y + 1. Then the set f = 0 is the line {(t,−1 − t) : t ∈ C}in C2, so is one-dimensional. The tropical variety, shown on the right in Figure 2, isalso one-dimensional.

It is true in general (we will see later) that dimension is preserved under tropical-ization. Other (primarily intersection theoretic) invariants are also preserved.

AARMS TROPICAL GEOMETRY 3

Motivating Example: Counting CurvesOne of the first successful applications of tropical geometry has been to enumerative

geometry, primarily in the work of Mikhalkin. This allows a simple answer to theclassical question of counting the number of rational curves in P2 of a given degree dpassing through a set of fixed points in general position. A curve C in P2 is given bya homogeneous polynomial f(x, y, z) = 0. The curve C is rational if it is isomorphicto P1 (informally, if there is a parameterization φ : C → C so C is the Zariski closureof the image of φ). The degree of the curve is the degree of the polynomial. In orderfor this number to be finite, we ask that the points be in general position, and thatthere be 3d−1 of them. Here “general position” means that there is a (Zariski) openset in (P2)3d−1/S3d−1 for which this number is constant.

Definition 2.1. LetN0,d be the number of rational curves of degree d passing througha collection of 3d− 1 points in P2 in general position.

Example:

d = 1 A curve of degree one is a straight line, which is rational. Thus N0,1 is one,as there is a unique line joining any two distinct points in P2.

d = 2 All curves of degree two are rational, and there is a unique curve through anyfive points in general position in P2 (see exercises). Thus N0,2 = 1.

d = 3 N0,3 = 12. This was computed by Steiner in 1848, and was possibly knownearlier.

d = 4 N0,4 = 620. This was computed by Zeuthen in 1873.d = 5 N0,5 = 87304. This (and all later ones) were unknown until the early 90s.d = 6 N0,6 = 26312976.

.In 1994 Kontsevich gave a recursive formula that determines all of these numbers

from N0,1 = 1. This involved developing the moduli space of stable maps, which isat the foundations of Gromov-Witten theory. Giving a self-contained proof of thiswould take more than this entire course. However in the last week we will (hopefully)give a self-contained proof of the Kontsevich recursion using tropical methods.

We now return to the question of tropicalizing polynomials. Earlier we saidtrop(x2 + y2 − 1) = min(2x, 2y, 0). The −1 turned mysteriously into a 0. We willnow partially explain this (though a full explanation will be later in the week).

Let

K = C{{t}} =⋃n≥1

C((t1/n)),

where by C((t1/n)) we mean the ring of Laurent series in the variable t1/n. This ringK is the ring of Puiseux series. An element a ∈ K has the form

a =∑q∈Q

aqtq,

where {q ∈ Q : aq 6= 0} is bounded below and has a common denominator.Write K∗ = K \ {0}. Let val : K∗ → R be given by val(a) = min{q : aq 6= 0}.

This lets us define the tropicalization of a polynomial formally.

4 DIANE MACLAGAN

Definition 2.2. Let S := C[x1, . . . , xn], and write

f =∑u∈Nn

cuxu,

where xu :=∏n

i=1 xuii . Then

trop(f) = minu∈Nn:cu 6=0

(val(cu) +

n∑i=1

uixi

).

The tropical hypersurface of f is

trop(V (f)) = {w ∈ Rn : the minimum in the definition of trop(f)(w) is achieved at least twice}.

Note that we could also define val : K → RC ∪ ∞ by setting val(0) = ∞. Thentrop(f) = minu∈Nn(val(cu)+

∑ni=1 uixi). Note also that val(−1) = 0, so this explains

the earlier zero.Example: Let f = tx2 + 2xy + 3ty2 + 5x + 7y − (t2 + t5). Then trop(f) =min(2x+ 1, x+ y, 2y + 1, x, y, 2). This function is illustrated in Figure 3.

(0, 0)(−1, 0)

(0,−1)

x

2x+ 1

y

2y + 1

x+ y

0

(2, 2)

Figure 3.

Example: Let f = (t2 − t5/2)y2 + 5x2 − 7xy + 8y − tx + t3. Then trop(f) =min(2 + 2y, 2x, x+ y, y, x+ 1, 3). This is illustrated in Figure 4.Example: Let f = tx2 + 3xy − 7(t3 + t5)y2 + ty − 7x + 5. Then trop(f) =min(2x+ 1, x+ y, 2y + 3, y + 1, x, 0). This is illustrated in Figure 5.Example: Let f = t2x− 7(t+ t3)y+ t5. Then trop(f) = min(x+ 2, y+ 1, 5). Thisis illustrated in Figure 6.

BExample: Let f = t3x2 − 7tx + 8xy − 7y2 + 6. Then trop(f) = min(2x + 3, x +1, x+ y, 2y, 0). This is illustrated in Figure 7.Example: Let f = t4x3 + t2x2y + t2xy2 + t4y3 + tx2 + xy + ty2 + x+ y + t. Thentrop(f) = min(3x+4, 2x+ y+2, x+2y+2, 3y+4, 2x+1, x+ y, 2y+1, x, y, 1). Thisis illustrated in Figure 8.

)


’

3x+ 1

2x

x+ y

y

2y + 2

(0,−2)

(0, 0)

(1, 2

(2, 3)

Figure 4.

0x

2x+ 1

x+ y

2y + 3

y + 1

(−1, 0)(0, 0

(1,−1)

(1,−2)

Figure 5.

Example: Let f = x2+2xy+3y2+4x+5y+6. Then trop(f) = min(2x, 2y, x, y, 0).This is illustrated in Figure 9.

Note two important aspects of these pictures: Firstly, in almost all cases, there arethe same number of “tentacles” in each of three directions, and that number is thedegree of the polynomial. Secondly, in almost all cases, at each vertex of the graph,

6 DIANE MACLAGAN

5

y + 1

x+ 2

(3, 4)

Figure 6.

0x+ 1

2x+ 3

2y

x+ y

(0, 0

(−1, 1)(−2, 1)

Figure 7.

the sum of the three vectors emanating from that vertex add to zero. In fact, withthe appropriate notion of multiplicity (see next week), both of these are true in allcases.

Outline of Course:Week 1: Introduction to varieties. Background tools, such as valuations and

Grobner bases.Week 2: Fundamental theorems on tropical varieties. Some basic examples.Week 3: More examples. Connections to toric varieties.Week 4: Enumerative geometry. Presentations.Presentations: During this month you will read a research paper or two in small

groups (3–5) and do a presentation on the material during the last few class periods.There is a list of possible papers on the webpage. Check that out today!!


(−2,−2)

x+ y

1x

y

2y + 1

3y + 4

2x+ 1

3x+ 4

2x+ y + 2

x+ 2y + 2

(0, 0)

Figure 8.

0

2x

2y

(0, 0)

Figure 9.

3. Lecture 2

The goal of today is to introduce affine and projective varieties. We adopt thesimplistic motto that “Algebraic geometry is the study of solutions of polynomialequations”. We are interested in the geometry of these solution spaces.

For example, consider the three equations x2+y2−1 = 0, xy = 0, and x2+y2 = −1.The first of these describes a circle of radius one, while the second is the union oftwo lines. The third has no solutions over the real numbers, but has solutions if workover the complex numbers (or at least an algebraically closed field), as we alwayswill.

8 DIANE MACLAGAN

Definition 3.1. Let k be an algebraically closed field (such as C). Then affine spaceis

An = {(a1, . . . , an) : ai ∈ k} = kn.

Philosophically An should be thought of as kn without a distinguished origin.

Definition 3.2. Let S = k[x1, . . . , xn] be the polynomial ring in n variables withcoefficients in k. Given f1, . . . , fs ∈ S the (affine) variety defined by the fi is

V (f1, . . . , fs) = {(a1, . . . , an) ∈ An : fi(a1, . . . , an) = 0 for 1 ≤ i ≤ s}.

Example: V (x+ y − 1) is the line y = x− 1.Example: V (x2−y, x3−z, y3−z2) = {(t, t2, t3) : t ∈ C)}. This is the affine “twistedcubic” curve.Note: V (f1, f2) = V (f1 + f2, f1 − f2) = V (f1, f2, f1 + f2, xf1 + y2f2).

Recall that an ideal I ⊆ S is a set closed under addition and multiplication byelements of S. The ideal I generated by f1, . . . , fs ∈ S is

I = 〈f1, . . . , fs〉 = {s∑i=1

gifi : gi ∈ S}.

Lemma 3.3. The variety V (f1, . . . , fs) only depends on the ideal 〈f1, . . . , fs〉, so if〈f1, . . . , fs〉 = 〈g1, . . . , gr〉 then V (f1, . . . , fs) = V (g1, . . . , gr).

Thus we will talk about varieties as defined by ideals. Note that if the ideal isprincipal (generated by one element) then we call the variety a hypersurface. All theexamples we saw yesterday were hypersurfaces.

Operations on varieties (The Ideal/Variety dictionary).

(1) V (I) ∩ V (J) = V (I + J). Here I + J = {f + g : f ∈ I, g ∈ J}. If I =〈f1, . . . , fs〉, J = 〈g1, . . . , gr〉, then I + J = 〈f1, . . . , fs, g1, . . . , gr〉.

(2) V (I) ∪ V (J) = V (I ∩ J) = V (IJ), where IJ = {fg : f ∈ I, g ∈ J}. IfI = 〈f1, . . . , fs〉, J = 〈g1, . . . , gr〉 then IJ = 〈figj : 1 ≤ i ≤ s, 1 ≤ j ≤ r〉. Adescription of I ∩ J in terms of the fi and gj is not as simple (though thereare algorithms to compute it).

Warning: We see here that we can have V (I) = V (J) if I 6= J (for exampleIJ 6= I ∩ J in general). For example,

V ((x− y)2) = V (x− y) = {(a, a) : a ∈ k}.

Solution: Hilbert’s Nullstellensatz.

Theorem 3.4. Let k be an algebraically closed field. Then V (I) = V (J) if and only

if√I =

√J , where √

I = {f ∈ S : f r ∈ I for some r}.

For a proof, see any book on commutative algebra (for example, [Eis95], or [CLO07]).

Example: If I = 〈x2〉, then√I = 〈x〉.

A subvariety of a variety V (I) is a variety V (J) with V (J) ⊂ V (I). Note that if

V (J) is a subvariety of V (I) then√I ⊆

√J .


We place a topology on An by setting the closed sets to be {V (I) : I is an ideal of S}.This is the Zariski topology. To check that ∅ and An are closed, note that ∅ = V (1),and An = V (0). Exercise: Check that the finite union of closed sets and the arbi-trary intersection of closed sets are closed. We denote by U the closure in the Zariskitopology of a set U . This is the smallest set of the form V (I) for some I that containsU .Another operation on varieties

(3) V (I) \ V (J) = V (I : J∞), where

I : J∞) = {f ∈ S : for all g ∈ J there exists N > 0 with fgN ∈ I}is the saturation of the ideal I by the ideal J .

Important (for us) example: Let J = 〈∏n

i=1 xi〉. Then V (J) = V (∏n

i=1 xi) =∪ni=1V (xi). For example, when n = 2, so S = C[x1, x2], then J = 〈x1x2〉, and V (J)is the union of the two coordinate axes. The complement An \ V (J) = T n = (C∗)n,

and for I ∈ S, V (I) \ V (J) = V (I) ∩ T n.Example: I = 〈x2

1+3x1x2〉, J = 〈x1x2〉. Then (I : J∞) = {f ∈ S : ∃N such that fxN1 xN2 ∈

I〉 = 〈x1 + 3x2〉.Definition 3.5. A variety X is irreducible if it cannot be written as the union oftwo proper subvarieties. This is a topological notion.

Proposition 3.6. Let X ⊂ An be a variety. Then X can be written uniquely (upto order) as an irredundant union of irreducible varieties. These are called the irre-ducible components of X.

Proof. We first show that such a decomposition exists. If X is irreducible then weare done. Otherwise we can write X = X1∪X2 where the Xi are proper subvarieties.Given a variety Y we write I(Y ) for the radical ideal defining Y . We must haveI(X) ( I(Xi) for i = 1, 2. Suppose now that we have a decomposition I = ∪si=1X

si .

If all of the Xi are irreducible, we are done. Otherwise there is some Xj that can bewritten in the form Xj = X ′

j ∪X ′′j where X ′

j, X′′j are proper subvarieties of Xj, so we

replaceXj byX ′j andX ′′

j in the decomposition and renumber to haveXs+11 , . . . , Xs+1

s+1 .

In this fashion we can get a decreasing sequence X1i1

) X2i2

) X3i3

) . . . s withcorresponding increasing sequence I(X1

i1) ( I(X2

i2) ( I(X3

i3) ( . . . . Since S is

Noetherian this sequence must terminate at some stage s, at which point each Xsi is

irreducible, and X = Xs1 ∪ · · · ∪Xs

s is an irreducible decomposition.Now suppose that X = X1 ∪ · · · ∪Xs = Y1 ∪ . . . Yr are two irredundant irreducible

decompositions of X. Since Y1 ⊂ X, ∪si=1(Y1 ∩ Xi) = Y1, so since Y1 is irreduciblethere must be 1 ≤ ji ≤ s with Y1 ⊂ Xji . Similarly for all other Yk there is jk withYk ⊂ Xjk . Reversing the roles of Xi and Yi, we also get for each 1 ≤ i ≤ s there is liwith Xi ⊆ Yli . But this means that Yi ⊆ Xji ⊆ Yk for k = lji , so we must have k = i,so for each 1 ≤ i ≤ r there is k with Xk = Yi. Since the decomposition in the Xi isirredundant, the Xi must be equal to the Yj up to order. �

Example: Let I = 〈x21 + 3x1x2〉. Then I = 〈x1〉 ∩ 〈x1 + 3x2〉, so V (I) = V (x1) ∪

V (x1 + 3x2). The varieties V (x1), V (x1 + 3x2) are both irreducible, so these are theirreducible components. The saturation of I by J = 〈x1x2〉 removes the componentV (x1) that does not intersect the torus.

10 DIANE MACLAGAN

Definition 3.7. The coordinate ring of X is S/I(X). This is the ring of polynomialfunctions on X.

Projective Varieties.

Definition 3.8. Projective space Pn = (Cn+1 \ 0)/ ∼ where v ∼ λv for all λ 6= 0.The points of Pn are the equivalence classes of lines through the origin 0. We write[x0 : x1 : · · · : xn] for the equivalence class of x = (x0, x1, . . . , xn) ∈ Cn+1. A line inPn is the equivalence class of a two-dimensional subspace in Cn+1.

We can think of Pn as An with some points “at infinity” added. For example, thereis a bijection between An and the points of Pn of the form [1 : x1 : · · · : xn]. The“points at infinity” are then those with first coordinate 0.Example: P2 = (C3 \ (0, 0, 0))/ ∼. Then P3 = {[1 : x1 : x2] : (x1, x2) ∈ A2} ∪ {[0 :x1 : x2] : x1, x2 ∈ C2} = {[1 : x1 : x2] : (x1, x2) ∈ A2} ∪ {[0 : 1 : x2] : x2 ∈ C} ∪ {[0 :0 : 1]}. So we see that the points at infinity are a copy of P1.Note: Polynomials don’t make sense as functions on Pn. For example, [1 : 2 : 3] =[2 : 4 : 6] ∈ P2, but the function x1 + x2 has different values (5 or 10) on these twopoints. However if f ∈ S is homogeneous, then {x ∈ Pn : f(x) = 0} is well-defined.This is because if f(x) = 0, then f(λx) = 0 for all λ 6= 0, since if f is homogeneousof degree k, then f(λx) = λkf(x).

We call an ideal I ⊂ k[x0, . . . , xn] homogeneous if it has a homogeneous generatingset.

Definition 3.9. Let I be a homogeneous ideal in S = k[x0, . . . , xn]. Then the varietyof I is

V (I) = {[x] ∈ Pn : f(x) = 0 ∀[x] ∈ Pn}.Example: V (x0 + x1 + x2) = {[1 : t : −1− t] : t ∈ C} ∪ {[0 : 1 : −1]}.Example: V (x0, x1, x2) = ∅.

The same rules apply for varieties in Pn as for An:

(1) V (I) ∩ V (J) = V (I + J);(2) V (I) ∪ V (J) = V (I ∩ J) = V (IJ);

(3) V (I) \ V (J) = V (I : J∞).

As in the affine case, there is not a bijection between homogeneous ideals andprojective varieties. Let m = 〈x0, . . . , xn〉. We call m the “irrelevant ideal”, as it isthe largest ideal not corresponding to a nonempty subvariety of Pn.Lemma 3.10. Let V (I), V (J) be subvarieties of Pn. Then V (I) = V (J) 6= ∅ if andonly if √

I =√J.

Also, V (I) = ∅ if and only if I = 〈1〉 or√I = m.

Proof. We first consider the case V (I) 6= ∅. Let V (I), V (J) denote the subvarieties

of An+1 defined by I and J . Note that if x ∈ V (I) then λx ∈ V (J) for all λ 6= 0 (and

similarly for V (J)). Thus V (I) = V (J) if and only if (V (J)) \ {0)} = (V (J)) \ {0)}.Now if f ∈ S satisfies f(λx) = 0 for all x ∈ V (I) \ 0 then f(0) = 0, so V (I) \ 0 =

V (I). Thus V (I) \ 0 = V (J) \ 0 if and only if√I =

√J by the Nullstellensatz.


Also V (I) = ∅ if and only if V (I) = ∅ or V (I) = {0}, so if and only if I = 〈1〉 or√I = m. �

Definition 3.11. The homogeneous coordinate ring of a projective varietyX = V (I)is S/I, where S = k[x0, . . . , xn].

Subvarieties of tori.The last case of varieties that we will consider is that of subvarieties of tori. This

is actually a special case of affine varieties. Let S = k[x±11 , . . . , x±1

n ] be the ring ofLaurent polynomials.Example: f = 3x1x

22 + 5x1x

32 + 7x−5

1 x2 ∈ S.The ring S is the coordinate ring of the algebraic torus T n ∼= (k∗)n. The name

comes from the fact that (C∗)n deformation retracts to the standard topologicaln-dimensional torus (S1)n. The ring S is the ring of all those rational functions(quotients of polynomials) that are defined everywhere on T n.

An ideal I ⊂ S determines a subvariety

V (I) = {x ∈ T n : f(x) = 0 for all f ∈ I} ⊆ T n.

Note that it makes sense to consider f(x) for x ∈ T n, since any f ∈ S has the formg/(∏n

i=1 xi)N for some polynomial g and N ≥ 0, so is defined at any x ∈ T n.

Note that a subvariety of T n is actually also an affine variety, which can be em-bedded into An+1. If X = V (I) ⊂ T n, choose a generating set for I consistingof polynomials {f1, . . . , fs}. This can always be done, since every monomial is aunit in S. Let S ′ be the polynomial ring k[x1, . . . , xn, y], and let J be the ideal〈f1, . . . , fs, y

∏ni=1 xi − 1〉, where we consider the fi here as elements of S ′. Then the

affine variety of J in An+1 consists of the points {(x, 1/∏n

i=1 xi) ∈ An+1 : x ∈ V (I) ⊂T n}.Warning: You’ll notice I’m using S for three different rings here: S = k[x1, . . . , xn],the coordinate ring of An; S = k[x0, . . . , xn], the homogeneous coordinate ring of Pn;and S = k[x±1

1 , . . . , x±1n ], the coordinate ring of T n. The meaning should always be

clear from context, and this has the advantage that one can summarize the previousdiscussion in the following form:

Note also that in each case there is a largest ideal defining a variety X, which wedenote by I(X). For example if X = V (I) ⊆ An, then I(X) =

√I.

Summary:Let X, Y be subvarieties of An, Pn or Tn, with ideals I(X), I(Y ) in the respective

coordinate ring. Then

(1) I(X ∪ Y ) = I(X) ∩ I(Y ) = I(X)I(Y );(2) I(X ∩ Y ) = I(X) + I(Y );

(3) I(X \ Y ) = I(X) : I(Y )∞;(4) The coordinate ring of X is S/I(X).

DimensionWe will study many invariants of a variety X. A basic one is the dimension of X.

We first give an intuitive definition of dimension. Nice (“smooth” or “nonsingular”)complex varieties are real manifolds of dimension 2d for some integer d. We say thatthe (complex) dimension of such an X is d.

12 DIANE MACLAGAN

Example: The projective variety P1 is equal to the two-dimensional sphere S2 asa set. This has real dimension two as a manifold, so the dimension of P1 is one.

Saying that dim(X) = d is intuitively saying that near most points X looks likeCd. (Intentionally vague sentence!)

Formally, the dimension of an irreducible variety X is the length d of the longestchain

∅ 6= X0 ( X1 ( · · · ( Xd = X

of irreducible subvarieties. (Note that this definition works for subvarieties of An,Pn, T n.)Example: {V (x1, x2) = (0, 0)} ( V (x1) ( A2, so dim(A2) ≥ 2. In fact dim(A2) = 2(and dim(An) = n for all n), but this is (surprisingly?) not trivial.Example: {(1, 1) = V (x1 − 1, x2 − 1) ( V (x2

1 + x22 − 1), so the dimension of

V (x21 + x2

2 − 1) is at least one. Again, in this case it is exactly one.There is an equivalent algebraic definition of dimension. The Krull dimension of

a ring R is the length d of the longest chain

P0 ( P1 ( · · · ( Pd = R

of prime ideals. If X ⊂ An or X ⊂ T n then dim(X) is the Krull dimension of thecoordinate ring S/I(X). If X ⊂ Pn then dim(X) is one less than the dimension ofS/I(X). For an overview of Krull dimension, see [Eis95, Chapter 8].

4. Exercises

These questions cover approximately Monday - Wednesday of week one. You donot need to do every question! This week some of you may have seen some of thecontent before, so concentrate on the new material. Do at least one question fromeach days material - ask me for advice on which questions are most appropriate foryour background if you’re not sure. You are strongly encouraged to work together.I will ask you to create a solution set as a group. This will involve typing up theanswer to approximately one question each a week.

Tropical Questions

(1) Check that (R,⊕,⊗) is a semiring.(2) Draw a picture of the tropical curve corresponding to the following polyno-

mials in K[x, y]:(a) f = t3x+ (t+ 3t2 + 5t4)y + t−2;(b) f = (t−1 + 1)x+ (t2 − 3t3)y + 5t4;(c) f = t3x2 + xy + ty2 + tx+ y + 1;(d) f = 4t4x2 + (3t+ t3)xy + (5 + t)y2 + 7x+ (−1 + t3)y + 4t;(e) f = tx2 + 4xy − 7y2 + 8;(f) f = t6x3 + x2y + xy2 + t6y3 + t3x2 + t−1xy + t3y2 + tx+ ty + 1.

(3) The goal of this exercise is to show the connection between tropical curvesin the plane and triangulations of a certain point configuration. It requiressome basic knowledge of polyhedral geometry (and is probably the hardestexercise in this problem set). Ask for hints/help once you’ve thought aboutit a little.


Fix d > 0. Let Ad = {(a, b) : a + b ≤ d, a, b ≥ 0}. Fix a polynomial f =∑(a,b)∈Ad

cabxayb with cab ∈ C{{t}}. The regular triangulation of Ad induced

by f is obtained by taking the convex hull of the points {(a, b, val(cab) : (a, b) ∈A} and taking the (projections of the) set of lower faces. These are the facesthat you can see if you look from (0, 0,−N) for N � 0.

Example: Let d = 2, so A2 = {(2, 0), (1, 1), (0, 2), (1, 0), (0, 1), (0, 0)}.Let f = tx2 + xy + ty2 + x + y + t6. We form the convex hull of the points{(2, 0, 1), (1, 1, 0), (0, 2, 1), (1, 0, 0), (0, 1, 0), (0, 0, 6)}. The lower faces of thispolytope are illustrated in Figure 10.

(0, 0)

(0, 2)

(1, 1)

(2, 0)

(0, 1)

(1, 0)

Figure 10.

(a) Draw the regular triangulation of A2 corresponding to the polynomialf = tx2 + xy + t3y2 + x+ ty + 1.

(b) Draw the regular triangulation of A1 corresponding to the polynomialf = t5x+ t3y + t10.

(c) Draw the regular triangulation of A3 corresponding to the polynomialf = t3x3 + tx2y + txy2 + t3y3 + tx2 + xy + ty2 + tx+ ty + t3.

The dual graph to a triangulation has a vertex for every triangle. Thereare two types of edges. The finite edges join two adjacent triangles, andhave direction orthogonal to the common edge of the triangles. The infiniteedges start at the triangles adjacent to the boundary of the large triangleconv((d, 0), (0, d), (0, 0)), and have direction orthogonal to the external edge.This is defined up to the lengths of the finite edges.

Example: In the example above, a dual graph for the regular triangula-tion is shown in Figure 11.(d) Draw a dual graph to the regular triangulation of A2 corresponding to

f = tx2 + xy + t3y2 + x+ ty + 1.(e) Draw a dual graph to the regular triangulation of A1 corresponding to

f = t5x+ t3y + t10.(f) Draw a dual graph to the regular triangulation of A3 corresponding to

f = t3x3 + tx2y + txy2 + t3y3 + tx2 + xy + ty2 + tx+ ty + t3.(g) Let f =

∑(a,b)∈Ad

cabxayb with cd0, c0d, c00 6= 0. Show that the tropical

curve defined by f is the image under x 7→ −x of a dual graph to theregular triangulation defined by f .

14 DIANE MACLAGAN

Figure 11.

(h) Check the previous claim for the examples of the first question.(i) Conclude that for sufficiently general f there are d tentacles pointing in

each direction. What can you say about the genericity condition? Whathappens in the other cases?

Varieties

(1) If you haven’t already done so, read a proof of the Nullstellensatz. Suggestedreferences: Cox, Little, O’Shea, or Eisenbud’s commutative algebra course.

(2) Show that if 〈f1, . . . , fs〉 = 〈g1, . . . , gr〉 then V (f1, . . . , fs) = V (g1, . . . , gr).(3) Show that V (I) ∩ V (J) = V (I + J).(4) Show that V (I) ∪ V (J) = V (IJ) = V (I ∩ J).

(5) Show that V (I) \ V (J) = V (I : J∞).

(6) Let I = 〈x2, xy3, y2z, z4〉 ⊂ k[x, y, z]. Compute√I. What are the irreducible

components of V (I)?(7) Show that the Zariski topology is a topology.(8) Describe the subvariety of P3 defined by the ideal I = 〈x0x2 − x1x3, x0x2 −

x21, x1x3 − x2

2〉. Repeat for I = 〈x22 − x1x3, x

21 − x0x2, x1x2x3 − x0x

23, x0x1x2 −

x20x3〉. Explain what you notice. (You may find a computer algebra system

helps here - ask around until you find a fellow student who knows how to useone if you don’t).

(9) Show that if X is an affine or projective variety or a subvariety of a torus, thenthere is a largest ideal I ⊂ S with X = V (I) in the sense that if X = V (J)then J ⊆ I.

(10) What is the dimension of the affine variety V (I) for I = 〈x1, x2〉 ⊂ A5? Whatabout the affine variety V (x2

1 − 3x2x3) ⊂ A3? What is the dimension of theprojective variety V (〈x0x2 − x1x3, x0x2 − x2

1, x1x3 − x22〉)?

(11) Let I = 〈x21 + x2, x

22 + x3〉 ⊂ k[x1, x2, x3]. What is the multiplicity of the

intersection of the affine varieties V (I) and V (xi) for i = 1, 2, 3?(12) Show that there is a unique curve of degree two through any five points in P2

in general position. What is the genericity condition?

5. Lecture 4

Today we will discuss valuations and Puiseux series.


Let K be a field. We denote by K∗ the nonzero elements of K. A valuation on Kis a function val : K → R ∪∞ satisfying

(1) val(a) = ∞ if and only if a = 0,(2) val(ab) = val(a) + val(b) and(3) val(a+ b) ≥ min{val(a), val(b)} for all a, b ∈ K∗.

We will always assume that 1 ∈ im(val). Since (λ val) : K → R is a valuation forany valuation val and λ ∈ R>0, this is not a serious restriction.Example: K = k(x), the ring of rational functions. We can write any functionf/g ∈ K as a Laurent series h =

∑hix

i where hi = 0 for i � 0. Then val(f/g) =min(i : hi 6= 0). If i is the lowest exponent occuring in f and j is the lowest exponentoccuring in g, then val(f/g) = i− j.Example: K = Q, and valp(q) = j when q = pja/b, where p does not divide a orb. For example val2(12/5) = 2, while val2(1/10) = −1. This the p-adic valuation.

Lemma 5.1. If val(a) 6= val(b) then val(a+ b) = min(val(a), val(b)).

Proof. Without loss of generality we may assume that val(b) > val(a). Since 12 = 1,we have val(1) = 0, and so (−1)2 = 1 implies val(−1) = 0 as well. Thus val(−b) =val(b), so val(a) ≥ min(val(a+ b), val(−b)) = min(val(a+ b), val(b)), and so val(a) ≥val(a + b). But val(a + b) ≥ min(val(a), val(b)) = val(a), and thus val(a + b) =val(a). �

Given a valuation val we define the valuation ring

R = {a ∈ K : val(a) ≥ 0} ∪ {0}.This is closed under addition and multiplication, since val(a), val(b) ≥ 0 impliesval(ab), val(a+ b) ≥ 0. It has a unique maximal ideal

m = {a ∈ K : val(a) > 0} ∪ {0}.To see that m is the unique maximal ideal, it suffices to note that if a ∈ R \m thena is a unit in R. Indeed, if a ∈ R \m, then val(a) = 0, so val(a−1) = − val(a) = 0, soa−1 ∈ R. The residue field is

k = R/m.

Example: If K = k((x)) is the quotient ring of k[[x]], then R = k[[x]], andR/m = k.Example: In the case that K = Q and val is the p-adic valuation, we haveR = {pja/b : j ≥ 0} ∪ {0}. Exercise: Check that the residue field is isomorphic toZ/pZ.Example: Let Rn = k[[t1/n]], and let k((t1/n)) be its quotient field. Let K =⋃n≥1 k((t1/n)), which we denote by k{{t}}. Note that K is closed under addition

and multiplication, and is thus a field. The field K is the ring of Puiseux series. Anelement of K has the form

∑q∈Q aqt

q where {q : aq 6= 0} is bounded below and hasa common denominator.

The field k((t1/n)) has a valuation like that on the ring of rational functions. Thisinduces a valuation val : K → R ∪∞. If a =

∑q∈Q aqt

q ∈ K, then val(a) = min{q :

aq 6= 0}.Puiseux series are useful because they are algebraically closed, as we now prove.

16 DIANE MACLAGAN

Theorem 5.2. If k is an algebraically closed field of characteristic zero, then K =k{{t}} is algebraically closed.

I learned the following proof from Thomas Markwig, and it is closely modelled onthe one he gives in his paper [Mar07] on a generalization of the Puiseux series.

Proof. We need to show that given a polynomial F =∑n

i=0 cixi ∈ S = K[x] there is

y ∈ K with F (y) =∑n

i=0 ciyi = 0. In principle the idea is to build y up as a Puiseux

series by successive powers of t.We first note that we may assume the following properties of F :

(1) val(ci) ≥ 0 for all i,(2) There is some j with val(cj) = 0,(3) c0 6= 0, and(4) val(c0) > 0.

To see this, note that if α = min{val(ci) : 0 ≤ i ≤ n} then multiplying F by t−α

does not change the existence of a root of F , which deals with the first two properties.If c0 = 0 then y = 0 is a root so there is nothing to prove.

To make the last assumption, suppose that F satisfies the first three assumptionsbut val(c0) = 0. If val(cn) > 0 then we can form G(x) = xnF (1/x) =

∑ni=0 cn−ix

i,which has the desired form, and if G(y′) = 0 then F (1/y′) = 0. If val(c0) = val(cn) =0 then consider the polynomial f := F ∈ k[x] that is the image of F in K[x]/mK[x].This which is not constant since val(cn) = 0. Since k is algebraically closed, we canchoose a root λ ∈ k of f . Then

F ′(x) := F (x+ λ) =n∑i=0

(n∑j=i

cj

(j

i

)λj−i)xi

has constant term F ′(0) = F (λ) with positive valuation, and F ′ still satisfies the firstthree properties. If y′ is a root of F ′, then y′ + λ is a root of F .

Set F0 = F . We will construct a sequence of polynomials Fi =∑n

j=0 cijxj. Suppose,

as we have shown we may assume for i = 0, that Fi satisfies conditions 1 to 4 above.The Newton polygon of Fi is the convex hull of the points {(i, j) : there is k with k ≤i, val(ck) ≤ j} ⊂ R2. There is an edge of the Newton polygon with negative slopeconnecting the vertex (0, val(ci0)) to a vertex (ki, val(ciki

)). Let

wi =val(ci0)− val(ciki

)

ki.

Let fi be the image in k[x] of the polynomial t− val(ci0)F (twix) ∈ K[x]. Note thatfi has degree ki, and has nonzero constant term. Since k is algebraically closedwe can find a root λi of fi. Let ri+1 be the multiplicity of λi as a root of fi, sofi = (x− λi)

ri+1gi(x), where gi(λi) 6= 0. Set

Fi+1(x) = t− val(ci0)Fi(twi(x+ λi)) =

n∑j=0

ci+1j xj.


Note that the coefficients ci+1j are given by the formula

(1) ci+1j =

n∑l=j

ciltlwi−val(ci0)

(l

j

)λl−ji .

The image of this in k is

ci+1j =

1

j!

∂jfi∂xj

(λi).

Note that we are using the characteristic zero assumption here. For 0 ≤ j < ri+1

this is zero, since λi is a root of fi of multiplicity ri+1. For j = ri+1 this is nonzero.Thus val(ci+1

j ) > 0 for 0 ≤ j ≤ ri+1, and val(ci+1j ) = 0 for j = ri+1. Note that we are

using the fact that char(k) = 0 here.If ci+1

0 = 0 then x = 0 is a root of Fi+1, so λitwi is root of Fi and so by recursing

we get∑i

j=0 λitw0+···+wj is a root of F0 = F , and we are done. Thus we may assume

that for each i we have ci+10 6= 0, so Fi+1 satisfies conditions 1 to 4 above, so we can

continue.The observation above on val(ci+1

j ) implies that ki+1 ≤ ri+1 ≤ ki. Since n is finite,the value of ki can only drop a finite number of times, so there is 1 ≤ k ≤ n andm for which for i ≥ m we have ki = k. This means that ri = k for all i > m, sofi = µi(x− λi)

k for all i > m, and some µi ∈ k.Let Ni be such that cij ∈ k((t1/Ni)) for 0 ≤ j ≤ n. We can take Ni+1 to be the

least common denominator of Ni and wi by Equation 1. Let yi =∑i

j=0 λitw0+···+wj ∈

k((t1/Ni)). We now show that we can take Ni+1 = Ni for i > m. In that case, wehave wi = val(ci0)/k, so it suffices to show that for i > m we have val(ci0) ∈ k/NiZ.This follows from the fact that fi is a pure power, so val(cij) = (k − j)/k val(ci0) for

1 ≤ j ≤ k, and in particular val(cik−1) = 1/k val(ci0) ∈ 1/NiZ. Thus there is an N for

which yi ∈ k((t1/N)) for all i, and so the limit

y =∑j≥0

λjtw0+···+wj ∈ k((t1/N)).

It remains to show that that y is a root of F . To see this, consider zi =∑

j≥i λjtwi+···+wj ,

and note that y = yi−1 + tw0+···+wi−1zi for i > 0, so

Fi(zi) = tval(ci0)Fi+1(zi+1).

Since z0 = y, it follows that

val(F (y)) =i∑

j=0

val(cj0) + val(Fi+1(zi+1)) >i∑

j=0

val(cj0)

for all i > 0. Since val(cj0) ∈ 1/NZ, we conclude that val(F (y)) = ∞, so F (y) = 0 asrequired. �

If k has characteristic p > 0 then k{{t}} is not algebraically closed. This is becausethe Artin-Schreier polynomial xp − x− t−1 has p “roots” of the form

(t−1/p + t−1/p2 + t−1/p3 + . . . ) + c

18 DIANE MACLAGAN

for c in the prime field Fp of k. These are not Puiseux series, since there is nocommon denominator of the exponents, but do live in the field of generalized powerseries which we now define.

Fix an algebraically closed field k, and a divisible group G ⊂ R. The Mal’cev-Neumann ring K = k((G)) of generalized power series is the set of formal sums α =∑

g∈G αgtg in an indeterminant t with the property that supp(α) := {g ∈ G : αg 6= 0}

is a well-ordered set.If β =

∑g∈G βgt

g then we set α+β =∑

g∈G(αg+βg)tg, and αβ =

∑h∈G(

∑g+g′=h αgβg′)t

h.

Then supp(α + β) ⊆ supp(α) ∪ supp(β), so is well-ordered, and thus α + β is well-defined. For αβ, define supp(α) + supp(β) to be the set {g + g′ : g ∈ supp(α), g′ ∈supp(β)}. Then supp(α) + supp(β) is well-ordered, and the set {(g, g′) : g + g′ = h}is finite for all h ∈ G, so multiplication is well-defined.

The field of generalized power series is the most general field with valuation weneed to consider in the following sense.

Theorem 5.3. Fix a divisible group G and a residue field k. Let K be a field witha valuation val with value group G such that val is trivial on the prime field (FP orQ) of K, and K has residue field k. Then K is isomorphic to a subfield of k((tG)).

One reference for these topics is [Poo93].

6. Lecture 5

The goal for today is to discuss Grobner bases in our contexts. From now onwe will always have K being an algebraically closed field with a nontrivial valuation(such as C{{t}}) with residue field k. We will discuss Grobner bases in three differentcontexts.

The homogeneous case. We first consider the case where there is an inclusion of kinto K with the image having valuation zero. This is the case for the Puiseux series,but not for all possible K. When an ideal has generators in k ⊂ K, we say that thecorresponding variety is defined over k, and that we are in the constant coefficientscase.

In this case, we first let S = k[x0, . . . , xn]. Fix w ∈ Rn. Given a polynomialf =

∑u∈Nn+1 cux

u ∈ S, set W = min{(0, w) · u : cu 6= 0}. Then

inw(f) =∑

(0,w)·u=W

cuxu.

If I is a homogeneous ideal, then we set

inw(I) = 〈inw(f) : f ∈ I〉.Example: Let f = x2

0 + 3x0x1. When w = 2, inw(f) = x20. When w = 0, inw(f) =

x20 + 3x0x1. When w = −3, inw(f) = 3x0x1. If I = 〈x2

0 + 3x0x1〉, then in2(I) = 〈x20〉.

Example: I = 〈x0x2−x21, x0x1−x2

2〉. Take w = (2, 3). Then inw(x0x2−x21) = x0x2,

and inw(x0x1 − x22) = x0x1. However inw(I) 6= 〈x0x2, x0x1〉, since x3

1 − x22 ∈ I, and

inw(x31 − x3

2) = x31. In this case inw(I) = 〈x0x2, x0x1, x

31〉.

Definition 6.1. A set {g1, . . . , gs} ⊂ I is a Grobner basis for w if inw(I) is generatedby {inw(g1), . . . , inw(gs)}.


Figure 12.

Example: With I as above, and w = (−1,−1), inw(I) = 〈x21, x

22〉. With w = (0, 0),

inw(I) = 〈x0x2 − x21, x0x1 − x2

2〉. With w = (1, 2), inw(x0x2 − x21) = x0x2 − x2

1, andinw(x0x1 − x2

2) = x0x1, and we have inw(I) = 〈x0x2 − x21, x0x1〉.

Definition 6.2. An ideal in S is monomial if it is generated by monomials. We saythat w is generic with respect to I if inw(I) is a monomial ideal.

Lemma 6.3. If w is generic then the monomials not in inw(I) form a k-basis forS/I.

We put an equivalence relation on Rn by setting w ∼ w′ if inw(I) = inw′(I).Example: With I as above, for w = (−2,−3), we have inw(I) = 〈x2

1, x22), so

(−1,−1) ∼ (−2,−3).

Theorem 6.4. The set

C[w] := {w′ ∈ Rn : inw(I) = inw′(I)}is a relatively open polyhedral cone. This means that it can be described by equationsand strict inequalities.

For a proof of this, see [Stu96, Chapter 2], or [Mac].Example: Let I be as above, and w = (−1,−1). Then

C[w] = {w′ ∈ R2 : 2w′1 < w′

2, 2w′1 < w′

2}.This is shown in Figure 12.

Definition 6.5. A polyhedral cone is the intersection of finitely many halfspaceswith the corresponding hyperplanes passing through the origin. It is thus of the form

C = {x : Ax ≤ 0}where A is a d × n matrix. A hyperplane H in Rn is supporting for a cone C if Clies in one of the two halfspaces determined by H. A face of C is the intersection ofC with a supporting hyperplane.

A fan is a collection of polyhedral cones, the intersection of any two of which is aface of each.

Theorem 6.6. For a fixed ideal I the collection {C[w] : w ∈ Rn} is a polyhedral fan.

20 DIANE MACLAGAN

Figure 13. Polyhedral fans

Figure 14. Not a polyhedral fan

Definition 6.7. This fan is called the Grobner fan of I.

Example: Let I = 〈x0x1 − x22, x0x2 − x2

1〉. Then the Grobner fan for I is shownin Figure 15, where the ideals corresponding to the cones are given in the followingtable.

Cone Initial ideal Cones Initial idealA 〈x0x1, x

21, x

20x2〉 a 〈x0x1, x0x2 − x2

1〉B 〈x0x2, x0x1, x

32〉 b 〈x0x2, x0x1, x

31 − x3

2〉C 〈x0x2, x0x1, x

31〉 c 〈x0x2, x0x1 − x2

2〉D 〈x0x2, x

22, x

20x1〉 d 〈x0x2, x

22, x

20x1 − x2

1x2〉E 〈x0x2, x

22, x

22x2, x

30x1〉 e 〈x0x2, x

22, x

21x2, x

30x1 − x4

1〉F 〈x0x2, x

22, x

21x2, x

41〉 f 〈x0x2 − x2

1, x22〉

G 〈x21, x

22〉 g 〈x0x1 − x2

2, x21〉

H 〈x0x1, x21, x1x

22, x

42〉 h 〈x0x1, x

21, x1x

22, x

30x2 − x4

2〉I 〈x0x1, x

21, x1x

22, x

30x2〉 i 〈x0x1, x

21, x

20x2 − x1x

21〉

There is software, called gfan [Jen], written by Anders Jensen, that will computethe Grobner fan of an ideal.


d

A

a

C

c

D

f

G

h

I

i

B b

H

g

F

eE

Figure 15.

The torus case. We now consider the case when S = k[x±11 , . . . , x±1

n ]. Given aLaurent polynomial f =

∑u∈Zn cux

u, and w ∈ Rn, we set W = min{w · u : cu 6= 0},and then

inw(f) =∑

w·u=W

cuxu.

If I is an ideal in S, then the initial ideal is

inw(I) = 〈inw(f) : f ∈ I〉.We make the same caveats as before on the fact that the initial ideal is not necessarilygenerated by the initial terms of generators.Example: Let f = x + 1 ∈ k[x±1], and let I = 〈x + 1〉. When w = 1, we haveinw(f) = 1, and inw(I) = 〈1〉. When w = −1, inw(f) = x and inw(I) = 〈x〉 = 〈1〉.Example: Let f = x + y + 1 ∈ k[x±1, y±1], and let I = 〈f〉. For w = (1, 1),inw(f) = 1, and inw(I) = 〈1〉. For w = (1, 0), inw(f) = y + 1, and inw(I) = 〈y + 1〉.If w = (1,−1) then inw(f) = y, and inw(I) = 〈y〉 = 〈1〉.

The Grobner fan does not exist in the same fashion, as we can see from this examplethat {w : inw(I) = 〈1〉} is not convex. However ignoring these cases gives a cone.The support of a polyhedral fan in Rn is the set of those w ∈ Rn lying in some cone ofthe fan. It follows from the following proposition that the set of w with inw(I) 6= 〈1〉has the support of a polyhedral fan.

If f =∑cux

u ∈ k[x1, . . . , xn], let W = max{|u| : cu 6= 0}, where |u| =∑n

i=1 ui.

The homogenization, f of f is f =∑cux

W−|u|0 xu ∈ k[x0, . . . , xn].

Proposition 6.8. Let I be an ideal in k[x±11 , . . . , x±1

n ]. Let I = I ∩k[x1, . . . , xn], and

let J ⊂ k[x0, . . . , xn] = 〈f : f ∈ I〉. Then

inw(J)|x0=1 = inw(I).

Proof. Exercise. �

Remark 6.9. The variety in Pn of the ideal J from Proposition 6.8 is the projectiveclosure of the variety of I in T n. This is the Zariski closure in Pn of the image of thevariety of I ⊂ T n under the map i : T n → Pn given by x 7→ (1 : x).

22 DIANE MACLAGAN

Corollary 6.10. Let I ⊂ k[x±11 , . . . , x±1

n ] and let J be the ideal defined in Proposi-tion 6.8. Then the support of the subfan of the Grobner fan of J consisting of thosecones σ for which (inw(J) : x∞0 ) 6= 〈1〉 for w ∈ σ is {w ∈ Rn : inw(I) 6= 〈1〉.Proof. By Proposition 6.8, we have inw(I) 6= 〈1〉 if and only if there is no power ofx0 in inw(J), and thus if and only if (inw(J) : x∞0 ) 6= 〈1〉. �

Nonconstant coefficients. We now consider the case that S = K[x±11 , . . . , x±1

n ].Fix w ∈ Rn, and let f =

∑u∈Zn cuxu ∈ S. Let W = min{val(cu) + w · u : cu 6= 0}.

Then

inw(f) = t−W∑u∈Zn

cutw·uxu ∈ k[x±11 , . . . , x±1

n ],

andinw(I) = 〈inw(f) : f ∈ I〉.

Example: Let f = (t + t2)x + t2y + t4, and w = (0, 0). Then W = 1, and

inw(f) = (1 + t)x = x. When w = (4, 2), W = 4, and inw(f) = y + 1. Whenw = (2, 1), W = 3, and inw(f) = x+ y.

Definition 6.11. A polyhedron is the intersection of finitely many (affine) halfspaces.Unlike a polyhedral cones the boundary (affine) hyperplanes are not required to passthrough the origin. An affine hyperplane H is supporting for a polyhedron P ifP ∩ H 6= ∅ and P lies on one side of H. A face of P is the intersection of P witha supporting hyperplane, or P itself. A polyhedral complex in Rn is a collection ofpolyhedra in Rn, the intersection of any two of which is a face of each.

As in the constant coefficient k case we can also consider initial ideal in the ho-mogenized polynomial ring K[x0, . . . , xn]. Instead of a Grobner fan, though, there isnow a Grobner complex. Each w ∈ Rn determines a relatively open polyhedron in Rn

on which the initial ideal is constant. Throwing away those polyhedra for which thecorresponding initial ideal contains a power of x0, we obtain the following theorem,whose proof we will omit.

Theorem 6.12. The set of w ∈ Rn for which inw(I) 6= 〈1〉 is the support of apolyhedral complex.

Example: Let f = tx2 + 2xy + 3ty2 + 4x + 5y + 6t ∈ C{{t}}[x±1, y±1], and letI = 〈f〉 be the ideal generated by I. Then the set {w ∈ R2 : 〈inw(I) 6= 〈1〉} isillustrated in Figure 16. The initial ideals corresponding to the labelled edges are aslisted in the following table.

Cone Initial idealA 〈x2 + 4x〉B 〈4x+ 6〉C 〈5y + 6〉D 〈4x+ 5y〉E 〈2xy + 4x〉F 〈x2 + 2xy〉G 〈2xy + 5y〉H 〈3y2 + 5y〉I 〈2xy + 3y2〉


I

A B

C

DE

F G H

Figure 16.

7. Exercises

Valuations

(1) Show that the residue field of k{{t}} is isomorphic to k.(2) Let K = Q with the p-adic valuation. Show that the residue field of K is

isomorphic to Z/pZ.(3) Show that if K is an algebraically closed field with a valuation val : K∗ → R,

and k = R/m is its residue field, then k is algebraically closed. Give anexample to show that if k is algebraically closed it does not automaticallyfollow that K is algebraically closed.

(4) In the proof that k{{t}} is algebraically closed, explain why fi has degree kiand has a nonzero constant term.

(5) Apply the algorithm implicit in the proof that C{{t}} is algebraically closedto compute (the start of) a solution to the equation x2 + t + 1 = 0. Checkyour answer with a computer algebra package (eg puiseux in maple).

Grobner bases

(1) Let I = 〈f〉 ⊂ k[x0, . . . , xn] be a principal ideal. Show that f is a Grobnerbasis for I.

(2) Compute all the initial ideals inw(f) of f = 7x20 + 8x0x1 − x2

1 + x0x2 + 3x22 as

w varies in R2. Draw the Grobner fan of 〈f〉. (Hint: start by choosing someparticular values of w).

(3) Show that if inw(I) is a monomial ideal for I ⊂ S = k[x0, . . . , xn] then themonomials not in inw(I) form a k-basis for S/I.

(4) In this question you will compute the Grobner fan of a principal ideal. TheNewton polytope of a polynomial f =

∑u∈Nn+1 cux

u ∈ k[x0, . . . , xn] is theconvex hull in Rn+1 of the exponents {u : cu 6= 0}.(a) Draw the Newton polytope of x2

0 + x0x1 + x21 + x2

2.If P is a polytope in Rn, a point v ∈ P is a vertex if there is w ∈ Rn for

which w ·v < w ·x for all x ∈ P \v. The normal cone to P at v is the closureof the set of all such w.

24 DIANE MACLAGAN

(b) Let P = conv((0, 0), (2, 0), (0, 2), (1, 1), (2, 2)). What are the vertices ofP? Draw the normal cone to each.

The normal fan of P is the union of the normal cones to vertices of P . Itis a polyhedral fan.(c) Draw the normal fan to the P of the previous question.(d) Show that the Grobner fan (as we have defined it) of 〈f〉 is the x0 = 0

slice of the normal fan to the Newton polytope of f .(5) (For people who already knew something about Grobner bases). It is more

standard to define an initial ideal using a term order on the polynomial ring.(a) Let f = x2

0 + x0x1 + x21 + x2

2. For each lexicographic or degree reverselexicographic term order ≺ find w ∈ R2 with inw(f) = in≺(f).

(b) In fact every term order can be represented by a vector w. You canread a proof, for example, in Proposition 2.4.4 of the notes available atwww.warwick.ac.uk/staff/D.Maclagan/papers/indialectures.pdf.gz

. See elsewhere in that chapter for hints on how to compute inw(I) usingyour favourite computer algebra package.

(6) Let f = t2x+ 3ty+ t4 ∈ K[x±1, y±1], where K = C{{t}}. Compute inw(f) forw = (2, 5), and w = (1, 2).

(7) Let f = x + y + 1. Draw {w ∈ R2 : inw(f) 6= 〈1〉}. Repeat this withf = tx2 + xy + ty2 + x + y + t. Compare your pictures with trop(V (f)) ineach case.

(8) Fix I ⊂ k[x±11 , . . . , x±1

n ]. Let I = I ∩ k[x1, . . . , xn], and let J = 〈f : f ∈I〉 ⊂ k[x0, . . . , xn], where f is the homogeneization of f using the variable x0.Show that

inw(J)|x0=1 = inw(I).

Optional extra: repeat with K.(9) (Open ended for the more computationally minded:) Play with the software

gfan (freely available fromhttp://www.math.tu-berlin.de/∼jensen/software/gfan/gfan.html).

(10) (Less open ended). If you don’t download gfan, find someone else in the classwho has.

8. Lectures 6 and 7

The goal for today is to define tropical varieties and state the fundamental theoremof tropical varieties.

As always, K is an algebraically closed field with a nontrivial valuation val : K →R ∪∞. It will never be wrong to take K = C{{t}}.

We first recall the definition of Grobner bases in S = k[x±11 , . . . , x±1

n ] from lasttime. Technically the definition I gave in the previous lecture used the notation t−W ,so only made sense in the Puisuex series field. We first define a notion of ta fora ∈ im(val) for an arbitrary algebraically closed field K with a valuation val.

Lemma 8.1. Let K be an algebraically closed field with a valuation val : K → R∪∞,and let im(val) be the additive subgroup of R that is the image of K∗ under val. Thesurjection of abelian groups K∗ � im(val) splits, so there is a group homomorphismφ : im(val) → K∗ with val(φ(w)) = w.


(−1, 0)

(1, 1)

(0, 0)

(0,−1)

Figure 17.

Proof. Since K is algebraically closed, it contains the nth roots of all of its elements.Thus K∗, and so im(val) are divisible abelian groups. Since im(val) is an additivesubgroup of R it is torsionfree, so im(val) is a torsionfree divisible group, and thusisomorphic to a (possibly uncountable) direct sum of copies of Q (see, for example,[Hun80, Exercise 8, p198]). Given any summand isomorphic to Q, with w ∈ im(val)taken to 1 by the isomorphism, and any a ∈ K∗ with val(a) = w, there is a homomor-phism φ : Q → K∗ taking w to a ∈ K∗. By construction this homomorphism satisfiesval(φ((m/n)w)) = (m/n)w. The universal property of the direct sum then impliesthe existence of a homomorphism im(val) → K∗ with the desired property. �

We use the notation tw to denote the element φ(w) ∈ K∗. We always assume1 ∈ im(val), and so N ⊆ im(val), so tn makes sense for any n ∈ N.

Fix w ∈ Rn. Given f =∑

u∈Zn cuxu ∈ S, let W = min{val(cu) + w · u : cu 6= 0}.

Then

inw(f) = t−W∑u∈Zn

cutw·uxu ∈ k[x±11 , . . . , x±1

n ],

and

inw(I) = 〈inw(f) : f ∈ I〉.Example: Let f = 3tx2+5xy+7ty2+9x+y+2t, and let I = 〈f〉 ⊂ C{{t}}[x±1, y±1].Fix w = (w1, w2) ∈ R2. Then

inw(f) = t−W (3t2w1+1x2 + 5tw1+w2xy + 7t2w2+1y2 + 9tw1x+ tw2y + 2t) ∈ C[x±1, y±1],

where

W = min(2w1 + 1, w1 + w2, 2w2 + 1, w1, w2, 1).

For example, if W = 2w1 + 1 and all other terms are larger, then inw(f) = 3x2, andinw(I) = 〈3x2〉 = 〈1〉. So for inw(I) 6= 〈1〉, a necessary condition is that the minimumin the definition of W is achieved twice!

For example, if 2w1 + 1 = w1 ≤ w1 + w2, 2w2 + 1, w2, 1, then w1 = −1, w2 ≥ 0. Inthis case inw(f) = 3x2 + 9x, so inw(I) = 〈3x2 + 9x〉 = 〈x + 3〉 6= 〈1〉. The set of wfor which inw(I) 6= 〈1〉 is illustrated in Figure 17.

26 DIANE MACLAGAN

(0, 0)

Figure 18.

We now recall from the first day of class the definition of the tropical hypersurface.Given f =

∑u∈Zn cux

u ∈ K[x±11 , . . . , x±1

n ] we defined

trop(f)(w) = min(val(cu) + w · u : cu 6= 0),

and

trop(V (f)) = {w ∈ Rn : the minimum in trop(f)(w) is achieved twice }.

Recall that if X ⊂ T n is a variety with (radical) ideal I then X =⋂f∈I{x ∈ T n :

f(x) = 0}.

Definition 8.2. Let X ⊆ T n be a subvariety of T n with (radical) ideal I. Then thetropical variety or tropicalization of X is

trop(X) =⋂f∈I

trop(V (f)).

Warning: In the “classical” world we have

X =⋂f∈G

V (f)

where G is any generating set for the ideal of X. The analogue is not true tropically.Example: Let X = V (x+y+1, x+2y+3) ⊆ T 2. Note that X = V (y+2, x−1) ={(1,−2)} ⊆ T 2. However trop(V (x+ y + 1)) = trop(V (x+ 2y + 3)) = {(u, v) ∈ R2 :u = v ≤ 0} ∪ {(u, v) ∈ R2 : u = 0 ≤ v} ∪ {(u, v) ∈ R2 : v = 0 ≤ u} as shown inFigure 18. Thus trop(V (x+ y+ 1))∩ trop(V (x+ 2y+ 3)) is the union of these threeline segments. However trop(V (y + 2)) is the line y = 0, while trop(V (x− 1)) is theline x = 0, so their intersection is the point (0, 0).


(2, 1)

Figure 19.

Fundamental Theorem of Tropical Geometry.

Theorem 8.3. Let X ⊆ T nK be a subvariety of T n with (radical) ideal I ⊆ K[x±11 , . . . , x±1

n ].Then the following subsets of Rn coincide:

(1) trop(X);(2) {w ∈ Rn : inw(I) 6= 〈1〉};(3) The closure in Rn of {(val(x1), . . . , val(xn)) ∈ Rn : x = (x1, . . . , xn) ∈ X}.

We will sketch a proof of this theorem below, starting in the hypersurface case(when I(X) is a principal ideal). We first illustrate the theorem with an example.Example: Let X = V (f) ⊆ T 2

K for f = tx + 3t2y + t3 ∈ K[x±1, y±1], and letI = 〈f〉. Here, as always if it is not otherwise indicated, we take K = C{{t}}. Thethree sets of Theorem 8.3 are constructed as follows.

(1) We have trop(f) = min(x+1, y+2, 3), so the set trop(V (f)) = {(u, v) ∈ R2 :u = 2 ≤ v} ∪ {(u, v) ∈ R2 : v = 1 ≤ u} ∪ (u, v) ∈ R2 : u = v+ 1 ≤ 2}. This isillustrated in Figure 19.

We are using here that the two possible definitions of trop(V (f)) coincide,so the set where the minimum in the definition of trop(f) is achieved twiceis equal to the intersection over all g ∈ I of the set where the minimum inthe definition of trop(g) is achieved twice. This is an exercise in the secondexercise set.

(2) Given w ∈ R2, the initial term inw(f) is the image in C[x±1, y±1] of

t−W (tw1+1x+ 3tw2+2y + t3),

where W = min(w1 +1, w2 +2, 3). If this minimum is achieved only once theninw(f) is a monomial, so inw(I) = 〈1〉. So if inw(I) 6= 〈1〉, then the minimumis achieved at least twice. Conversely, if the minimum is achieved at least

28 DIANE MACLAGAN

twice, then inw(f) is not a monomial. It is

x+ 3y if w1 + 1 = w2 + 2 < 3;

x+ 1 if w1 + 1 = 3 < w2 + 2;

3y + 1 if w2 + 2 = 3 < w1 + 1;

x+ 3y + 1 if (w1, w2) = (2, 1).

Thus, since inw(I) = 〈inw(f)〉, we have

{w ∈ R2 : inw(f) 6= 〈1〉} = trop(X).

(3) The variety X is

X = {(x, y) ∈ T 2K : tx+ 3t2y + t3 = 0}

= {(−t2 − 3ty, y) : y ∈ K∗, 3ty + t2 6= 0}.

Thus

{(val(x), val(y)) : (x, y) ∈ X}

= {(val(−t2 − 3ty), val(y)) : y ∈ K∗, y 6= −t/3}.

Now val(−t2−3ty) = min(2, val(y)+1) if y 6= −t/3+z with val(z) > 1. Thus

(val(x), val(y)) : (x, y) ∈ X} = {(2, w) : w ≥ 1}∪{(w+1, w) : w ≤ 1}∪{(w, 1) : w ≥ 2}.So all three sets coincide in this example.

We now begin the proof of Theorem 8.3. We first prove this in the hypersurfacecase.

Proposition 8.4. Let K be an algebraically closed field with a nontrivial valuationval, and let f ∈ K[x±1

1 , . . . , x±1n ]. Then the following three sets coincide.

(1) trop(V (f));(2) The set {w ∈ Rn : inw(f) is not a monomial }.(3) The closure of the set {(val(v1), . . . , val(vn)) : v ∈ T nK , f(v) = 0};

Proof. Let (w1, . . . , wn) ∈ trop(V (f)). Then by definition the minimum W =min(val(cu) + u · w : cu 6= 0) is achieved at least twice. This then means thatinw(f) =

∑u:cu 6=0,val(cu)+u·w=W cux

u is not a monomial, and thus set (1) is contained

in set (2). Conversely, if inw(f) is not a monomial, then min(val(cu) + u ·w : cu 6= 0)is achieved at least twice, so w ∈ trop(V (f)). This shows the other containment, sothe first two sets are equal.

We now prove the inclusion of set (3) in set (1). Since set (1) is closed, it isenough to consider points in (3) of the form val(v) := (val(v1), . . . , val(vn)) where v =(v1, . . . , vn) ∈ T nK satisfies f(v) = 0. Let v ∈ T nK satisfy f(v) = 0, so

∑u∈Zn cuv

u = 0.We first reduce to the case where val(cuv

u)) ≥ 0 for all u, so cuvu ∈ R. Let

W = min{val(cuvu) : cu 6= 0}, and let g = t−Wf . Then g(v) = 0, and trop(V (g)) =

trop(V (f)), so it suffices to prove the inclusion with f replaced by g. We can thus


assume (?) that val(cuvu) ≥ 0, and that there is at least one u with val(cuv

u) = 0.Then f(v) =

∑cuv

u = 0 is the sum of elements of R, and so we can consider theirimage in the residue field k = R/m. This is

∑cuvu = 0 ∈ k. By assumption

(?) at least one of the terms vu is nonzero. Since the sum of all such terms is0 ∈ k, we conclude that there must in fact be at least two terms with val(cuv

u) = 0.We have val(cuv

u) = val(cu) +∑ui val(vi) ≥ 0 for all u by assumption ?, so this

means that the minimum 0 = mincu 6=0(val(cu) + u · val(v)) is achieved twice, whereval(v) = (val(v1), . . . , val(vn)). Thus val(v) ∈ trop(V (f)) as required.

Finally, we prove the inclusion of set (1) into set (3). Since the image of thevaluation val is dense in R (see Exercises), and the set (3) is closed by definition,it suffices to consider a point in (1) of the form w = val(y) for some y ∈ T nK . Wewant to construct β = (β1, . . . , βn) ∈ T nK with val(β) = (val(β1), . . . , val(βn)) = wand f(β) = 0.

Let inw(f) =∑aux

u ∈ k[x±11 , . . . , x±1

n ]. Since inw(f) is not a monomial, there issome variable xi that appears to different powers in those xu with au 6= 0. Afterreordering, we can assume that this is x1. Replacing f by xm1 f for some m ∈ Z doesnot change trop(V (f)) or whether inw(f) is a monomial, so we may assume that x1

appears in some, but not all, of the monomials occuring in inw(f), and that if x1

occurs, the exponent is positive.Since k is algebraically closed and inw(f) is not a monomial, we can find α ∈ (k∗)n

with inw(f)(α) = 0. Let βi = αitwi for 2 ≤ i ≤ n. Let g(y) = f(y, β2, . . . , βn) ∈

K[y±1]. The assumption that the exponent of x1 in every monomial of f is non-negative and that some such exponents are zero means that in fact g ∈ K[y] is apolynomial of positive degree with nonzero constant term. Since K is algebraicallyclosed we can thus factor g into linear factors:

g = λm∏i=1

(y − bi).

Write u′ = (u2, . . . , un) ∈ Zn−1 for the projection of u onto the last n − 1 com-ponents. Then g(y) =

∑u∈Zn(cuβ

u′)yu1 . Note that val(βu′) =∑n

i=2 ui val(βi) =∑ni=2 uiwi, so val(cuβ

u′) + w1u1 = val(cu) + w · u.

Thus inw1(g)(y) =∑

u:val(cu)+w·u=W tw1u1−W cuβu′yu1 =∑

u:val(cu)+w·u=W auαu′yu1 ,

and so inw1(g)(α1) =∑auα

u = 0.

Now inw1(g) = t− val(λ)λ inw(y − b1) · · · inw(y − bm) (see Exercises). Thus there issome j for which inw(y − bj)(α1) = 0. For this j we must have val(bj) = w1, asotherwise inw(y − bj) is a monomial, and so then inw(y − bj)(α1) 6= 0 (since α1 6= 0because g has a nonzero constant term). Let β1 = bj. Then g(β1) = 0 by construction.Thus if β = (β1, β2, . . . , βn), we have f(β) = 0 and val(β) = w, so β is the desiredpoint. �

The proof of Theorem 8.3 in the general (non-hypersurface) case proceeds by re-duction to the hypersurface case, as we now sketch.

Sketch of proof of Theorem 8.3. The equality of the first two sets is immediate fromProposition 8.4, as w ∈ trop(X) = ∩f∈I trop(f) if and only if inw(f) is not a mono-mial for all f ∈ I, which occurs if and only if inw(I) 6= 〈1〉.

30 DIANE MACLAGAN

The inclusion of the third set into the first is also an immediate corollary of Propo-sition 8.4. If w = val(y) for y ∈ X, then f(y) = 0 for all f ∈ I, so w ∈ trop(V (f))for all f ∈ I, so w ∈ trop(X).

We are left with showing that the second set is contained in the third set. Thekey idea is to project X to a hypersurface. We can (proof skipped) assume that X isirreducible of dimension d. We then claim (proof skipped) that for a generic choiceof projection φ : T n → T d+1 the image φ(X) is a hypersurface in T d+1. Applying themap Hom(K∗,−) to the projection φ : T n → T d+1 we get a map ψ : Zn → Zd+1. Wewrite φ for both the projection T nK → T d+1

K and for the corresponding projection T nk →T d+1

k , and also for the maps of coordinate rings k[y±11 , . . . , y±1

d+1] → k[x±11 , . . . , x±1

n ].For a generic projection we have

φ(inw(I))) = inψ(w)(φ(I)).

Thus if w lies in the second set of the Theorem, so inw(I) 6= 〈1〉, then φ(inw(I)) =inψ(w)(φ(I)) 6= 〈1〉 ⊆ k[y±1

1 , . . . , y±1d+1]. Applying Proposition 8.4 we see that ψ(w) ∈

trop(φ(X)), so there is y ∈ φ(X) with val(y) = ψ(w). Since y ∈ φ(X) there is y ∈ Xwith φ(y) = y, and ψ(val(y)) = ψ(w). We claim (proof skipped) that we can choosey with val(y) = w, which shows that w lies in the third set of the theorem. �

Remark 8.5. Theorem 8.3 says that for a dense set of w ∈ trop(X) then there isy ∈ X with val(y) = w. Sam Payne has shown that the set

{y ∈ X : val(y) = w}is Zariski dense in X for all w ∈ trop(X). See [Pay07] for details.

We finish this lecture with examples of tropical varieties for which the classicalvariety is not a hypersurface.Example: Let X = V (x1 + x2 + x3 + 1, x2 + 2x3 + 3〉 ⊂ T 3. Then

X = V (x1 − x3 − 2, x2 + 2x3 + 3)

= {(2 + s,−3− 2s, s) : s ∈ K∗ : t 6= −2,−3/2}.

Now

val(2+,−3− 2s, s) =

(0, 0, val(s)) if val(s) > 0(val(s), val(s), val(s)) if val(s) < 0(w, 0, 0) if s = −2 + s′, val(s′) = w > 0(0, w, 0) if s = −3/2 + s′, val(s′) = w > 0(0, 0, 0) if val(s) = 0, s 6= −2,−3/2

Thus trop(X) is the union of the rays through (1, 0, 0), (0, 1, 0), (0, 0, 1), and (−1,−1,−1)in R3.Example: Let I = 〈x1 + x2 + x3 + x4 + 1, x2 + 2x3 + 3x4 + 4〉 ⊆ k[x±1

1 , . . . , x±14 ],

and let X = V (I) ⊆ T 4K . Then trop(X) is the two-dimensional fan in R4 with

{(1, 0, 0, 0), (0, 1, 0, 0, ), (0, 0, 1, 0), (0, 0, 0, 1), (−1,−1,−1,−1)}and two-dimensional cones spanned by any two of these. The intersection of trop(X)with the sphere S3 ⊆ R4 is then a graph with five vertices and ten edges (the completegraph K5).


It is hard to draw pictures of tropical varieties that do not lie in R2. For two-dimensional tropical varieties we will often resort to this trick of intersecting withthe sphere and drawing the corresponding graph.

9. Exercises

(1) Let K be an algebraically closed field with a nontrivial valuation val : K →R ∪∞. Show that im(val) is dense in R.

(2) Let f ∈ K[x±11 , . . . , x±1

n ]. Show that inw(fg) = inw(f) inw(g).(3) Let I ⊆ K[x±1

1 , . . . , x±1n ]. Show that if g ∈ inw(I) then g = inw(f) for some

f ∈ I.(4) Let f ∈ K[x±1

1 , . . . , x±1n ], and let I = 〈f〉. Show that trop(V (f)) = ∩g∈I trop(V (g)).

This can be rephrased as “hypersurfaces tropicalize to hypersurfaces”.(5) Let f = tx2

1 + x1x2 + tx22 + x0x1 + x0x2 + t4x2

0 ∈ C{{t}}[x0, x1, x2]. Computethe Grobner complex of I = 〈f〉 (ie compute the polyhedra on which inw(f)is constant). (Part of this exercise is taking the common generalization ofGrobner bases in C[x0, x1, x2] and those in C{{t}}[x±1

1 , x±12 ]). Use your answer

to draw the tropical variety of V (f) ⊆ C{{t}}[x±11 , x±1

2 ].(6) Verify (as much as possible) the fundamental theorem of tropical geometry

for X = V (f) for the following polynomials f ∈ C{{t}}[x±11 , x±1

2 ]:(a) f = 3x1 + t2x2 + 2t;(b) f = tx2

1 + x1x2 + tx22 + x1 + x2 + t;

(c) f = x31 + x3

2 + 1.(7) Let S = K[x±1

1 , . . . , x±14 ]. Describe trop(X) for the following subvarieties

of T 4. Hint: Both are two-dimensional, so you could draw the graph oftrop(X) ∩ S3.(a) X = V (x1 + x2 + x3 + x4 + 1, x2 + 2x3 + 3x4 + 4);(b) X = V (x1 + x2 + x3 + x4 + 1, x2 + x3 + 2x4 + 2).

(8) Let φ : T 2 → T 4 be given by

φ(t1, t2) = (t1, t1t2, t1t22, t1t

32).

Let X = im(φ). Compute trop(X).

10. Lecture 8

The goal for today is to start describing the structure of the tropical variety.Example: Let f = x + y + 1 ∈ K[x±1, y±1]. Then V (f) is a line in T 2, andtrop(V (f)) is the standard “tropical line” we have seen multiple times, as shown inFigure 20

32 DIANE MACLAGAN

(0, 0)

Figure 20.

Example: Let f = x + y + z + 1 ∈ K[x±1, y±1, z±1]. Then V (f) is a surface in T 3.We have w ∈ trop(V (f)) if and only if

w1 = w2 ≤ w3, 0

or w1 = w3 ≤ w2, 0

or w1 = 0 ≤ w2, w3

or w2 = w3 ≤ w1, 0

or w2 = 0 ≤ w1, w3

or w3 = 0 ≤ w1, w2.

This is a fan with rays 1

00

,

010

,

001

,

−1−1−1

.

The fan consists of all two-dimensional cones in R3 generated by any two of theserays. It intersects the sphere S3 in the complete graph K4.Example: Let φ : T d → T n be a subtorus embedded by

φ : s = (s1, . . . , sd) 7→ (sa1 , . . . , san),

where aj ∈ Zd for 1 ≤ i ≤ n, and saj =∏d

i=1 saij

i . We assume that the d× n matrixA = (aij) has rank d, so that φ is an embedding. Let X = im(φ) ∼= T d ⊂ T n. Then

trop(X) = closure of {val(sa1), . . . , val(san) : s = (s1, . . . , sd) ∈ T dK}= closure of {a1 · val(s), . . . , an · val(s) : s ∈ T dK}= closure of {AT val(s) : s ∈ T dK}= imAT .

So trop(X) is a linear space of dimension d.

Definition 10.1. The Minkowski sum of two subsets A,B ⊂ Rn is the set

A+B = {a+ b : a ∈ A, b ∈ B}.

Definition 10.2. The affine span of a polyhedron P ⊆ Rn is

aff(P ) = v + span(u− v : u ∈ P )


Figure 21. The complex on the left is pure, while the one on theright is not.

(−1,−1)

(0, 0) (1, 0)

(2,−1)

Figure 22.

where v ∈ P . Here the sum is an example of Minkowski addition. Note that this isindependent of the choice of v ∈ P . The relative interior of P is the interior of Pinside its affine span.

Definition 10.3. The dimension of a polyhedron P is the dimension of its affinespan. A polyhedral complex Σ is pure of dimension d if all maximal polyhedra in Σare d-dimensional.

Note: In each of the examples, trop(X) is a pure polyhedral complex and dim(X) =dim(trop(X)). This is true in general.

Recall that the support of a polyhedral complex in Rn is the subset of Rn obtainedby taking the union of all the polyhedra in the complex.

Theorem 10.4. Let X ⊂ T nK be an irreducible variety of dimension d defined by theprime ideal I ⊂ K[x±1

1 , . . . , x±1n ]. There is a polyhedral complex Σ that is pure of

dimension d whose support is trop(X).

The existence of the polyhedral complex we saw already in the discussion of theGrobner complex last week. The new material here is that this complex is pure ofdimension d.

To prove this we need some more Grobner basics, which we will see first in anexample.Example: Let f = tx2y+x2 +xy+ t2y+x+ t ∈ C{{t}}[x±1, y±1]. Then trop(f) =min(2x+ y + 1, 2x, x+ y, y + 2, x, 1), and trop(V (f)) is shown in Figure 22.

34 DIANE MACLAGAN

For w = (1, 0), we have inw(f) = xy+x+1 ∈ C[x±1, y±1]. Then trop(V (inw(f))) ={v ∈ R2 : 〈inv(inw(f))〉 6= 〈1〉} is shown in Figure 23.

Figure 23.

For w = (0, 0) we have inw(f) = x2 + xy + x, and trop(inw(f)) = {v ∈ R2 :〈inv(inw(f))〉 6= 〈1〉} is shown in Figure 24.

Figure 24.

If w = (1/2, 0), then inw(f) = x+ xy, and trop(V (inw(f))) is the x-axis {(x, y) ∈R2 : y = 0}.

If w = (1, 1), then inw(f) = x+ 1, and trop(V (inw(f))) is the y-axis {(x, y) ∈ R2 :x = 0}.

Note that in all cases the set trop(V (inw(f))) looks like the piece of trop(V (f))“near” the polyhedron containing w. More formally, it is the star, which we nowdefine, of the polyhedron containing w in the polyhedral complex trop(V (f)).

Definition 10.5. Let Σ be a polyhedral complex, and let σ ∈ Σ be a polyhedron.The star starΣ(σ) of σ ∈ Σ is a fan in Rn whose cones are indexed by those τ ∈ Σfor which σ is a face of τ . Fix w ∈ σ. Then the cone indexed by τ is the Minkowskisum

τ = {v ∈ Rn : ∃ε > 0 with w + εv ∈ τ}+ aff(σ)− w.

Example: For the polyhedral complex Σ shown on the left of Figure 25, the affinespan of the vertex σ1 is just the vertex itself. The star is the standard tropical lineshown on the right. For σ2 the affine span is the entire y-axis, and this is also thestar.

Lemma 10.6. Let Σ be a polyhedral complex in Rn, and σ ∈ Σ. Fix w in the relativeinterior of σ. Then

starΣ(σ) = {v ∈ Rn : w + εv ∈ Σ for sufficiently small ε > 0}.


σ2

σ1

star(σ1)

star(σ2)

Figure 25.

Definition 10.7. A subspace V ⊆ Rn is the lineality space of a polyhedron P ⊆ Rn

ifx ∈ P implies x+ v ∈ P for all v ∈ V.

If V is the lineality space of a polyhedron P then we often consider P/V in Rn/Vfor ease of visualization.

Note: The affine span aff(σ) of a polyhedron σ ∈ Σ lies in the lineality space ofevery cone in the fan star(σ).

Lemma 10.8. Let I ⊆ K[x±11 . . . , x±1

n ], and fix w, v ∈ Rn. Then there is ε > 0 suchthat

inv(inw(I)) = inw+εv(I).

Proof. We first note that it suffices to check that for all f ∈ K[x±11 , . . . , x±1

n ] there isε > 0 such that

inv(inw(f)) = inw+ε′v(f)

for all ε′ < ε.To see this, note that inv(inw(I)) is finitely generated by g1, . . . , gs ∈ k[x±1

1 , . . . , x±1n ]

and each generator gi is of the form inv(inw(fi)) for some fi ∈ I, so we can chooseε to be the minimum of the εi corresponding to these generating fi. Then gi =inv(inw(fi)) = inw+εv(fi), so inv(inw(I)) ⊆ inw+εv(I). Equality follows from the factthat we cannot have a proper inclusion of initial ideals.

We now prove the claim lemma for an individual polynomial. Let f =∑

u∈Zn cuxu.

Theninw(f) =

∑u∈Zn

cutw·u−Wxu,

where W = min(val(cu) + w · u : cu 6= 0) = trop(f)(w). Let W ′ = min(v · u :val(cu) + w · u = W ). Then

inv(inw(f)) =∑

v·u=W ′

cutw·u−Wxu.

Let δ = min(val(cu)+w ·u−W : val(cu)+w ·u > W ), and let M = max(v ·u : cu 6= 0).Set ε = δ/2M , and W ′′ = min(val(cu) + (w + εv) · u. Then by construction we have

W ′′ = W + εW ′

36 DIANE MACLAGAN

and

{u : val(cu) + (w + εv) · u = W ′′} = {u : val(cu) + w · u = W, v · u = W ′}.Thus inw+εv(f) = inv(inw(f)). �

Recall that we say that v ∈ Rn is generic for I if inv(I) is a monomial ideal. Thefollowing corollary allows us to compute Grobner bases with respect to nongenericweight vectors using standard computer algebra packages.

Corollary 10.9. Let I ⊂ K[x±11 , . . . , x±1

n ], and w ∈ Rn. Choose a vector v ∈ Rn

that is generic for inw(I). Then a Grobner basis G for I with respect to w + εv forsufficiently small ε is a Grobner basis for I with respect to w, and inw(I) = 〈inw(g) :g ∈ G〉.

Proof. Fix ε > 0 such that inw+εv(I) = inv(inw(I)), the existence of which is guaran-teed by Lemma 10.8. Let G = {g1, . . . , gr} be a Grobner basis for I with respect toinw+εv. Thus inw+εv(I) = 〈inw+εv(g1), . . . , inw+εv(gr)〉. The choice of ε was made toguarantee that inw+εv(gi) = inv(inv(gi)) for all i, so inw+εv(I) = 〈inv(inw(g1)), . . . , inv(inw(gr))〉 =inv(inw(I)), so {inw(g1), . . . , in( gr)} is a Grobner basis for inw(I) with respect to v,and thus inw(I) = 〈inw(g1), . . . , inw(gr)〉. �

Corollary 10.10. Let X ⊂ T nK, with X = V (I) for I ⊆ K[x±11 , . . . , x±1

n ] and let Σbe a polyhedral complex whose support is trop(X) ⊂ Rn. Fix w ∈ trop(X), and let σbe the polyhedron of Σ containing w in its relative interior. Then

trop(V (inw(I))) = starΣ(σ).

Proof. We have

trop(V (inw(I))) = {v ∈ Rn : inv(inw(I)) 6= 〈1〉}= {v ∈ Rn : inw+εv(I) 6= 〈1〉 for sufficiently small ε > 0}= {v ∈ Rn : w + εv ∈ trop(X) for sufficiently small ε > 0}= starΣ(σ),

where the last equality is by Lemma 10.6. �

To prove Theorem 10.4 we need the following Lemma.

Lemma 10.11. Let Y ⊆ T nk be equidimensional of dimension d (all irreducible com-ponents have the same dimension). Suppose that trop(Y ) is a linear subspace of Rn.Then there is a d-dimensional subtorus T ⊂ T nk such that V (I) consists of finitelymany T -orbits.

Proof. For a proof, see Lemma 9.9 of [Stu02] �

Proof of Theorem 10.4. Let Σ be a polyhedral complex with support trop(X), andlet σ be maximal polyhedron in Σ (so σ is not a proper face of any polyhedron inΣ). We need to show that dim(σ) = d. Fix w in the relative interior of σ. ByCorollary 10.10 we have trop(inw(I)) = starΣ(σ). Since σ is maximal, we have thatstarΣ(σ) = aff(σ) is a dim(σ)-dimensional linear subspace. Since I is prime, it followsfrom a resultof Kalkbrener and Sturmfels [KS95] that V (inw(I)) is equidimensional,


(−1, 0)

(1, 1)

(0, 0)

Figure 26.

so all irreducible components have the same dimension. By Lemma 10.11 it followsthat there is a subtorus T ⊂ T nK of dimension dim(σ) for which V (inw(I)) is theunion of finitely many T -orbits. Since dim(V (inw(I))) = dim(I) = d (Exercise!), itfollows that dim(σ) = d. �

11. Lecture 9

In this lecture we discuss another property of tropical varieties: that they areweighted balanced polyhedral complexes.Example: Consider the vertex (0, 0) of trop(V (x + y + 1)). There are three raysleaving (0, 0): pos((1, 0)), pos((0, 1)), pos((−1,−1)). Note that we have(

10

)+

(01

)+

(−1−1

)=

(00

).

Example: Let X = V (tx2 + x+ y + xy + t) ⊂ T 2K for K = C{{t}}. Then trop(X)

is shown in Figure 26.Then the star of the vertex (1, 1) has rays spanned by (1, 0), (0, 1), and (−1− 1),

which add to (0, 0). This is also the star of the vertex (−1, 0). At the vertex (0, 0)the star has rays (1, 1), (−1, 0), and (0,−1), which add to (0, 0).Example: Let X = V (x2 + xy + ty + 1) ⊂ T 2

K for K = C{{t}}. Then trop(X) isshown in Figure 27.

The star of the vertex (1,−1) has rays spanned by (1, 0), (0,−1), and (−1, 1),which add to zero. For the vertex (0, 0) the star has rays (1,−1), (−1,−1), and(0, 1). In this case(

1−1

)+

(−1−1

)+

(01

)=

(0

−1

)6=(

00

).

However,

(2)

(1

−1

)+

(−1−1

)+ 2

(01

)=

(00

).

We will define a notion of multiplicity on the top-dimensional polyhedra in trop(X)so that the corresponding sum is the zero vector. We review some commutative

38 DIANE MACLAGAN

(1,−1)

(0, 0)

Figure 27.

xy + 1

(1,−1)

y + 1

xy + y

x2 + 1

(0, 0)

x2 + xy

Figure 28.

algebra to give the precise definition, and then give the intuitive definition as givenin class.

Definition 11.1. Let S = K[x±11 , . . . , x±1

n ]. The localization of S at a prime idealP is the ring with elements {f/g : f, g ∈ S, g 6∈ P}. Given an ideal I ⊂ S, themultiplicity of P in I is mult(P, S/I) = dimK(SP/SP I). If I is radical, this is one ifV (P ) is an irreducible component of V (I), and zero otherwise.

Definition 11.2. Let X = V (I) be an irreducible variety of dimension d for I ⊂K[x±1

1 , . . . , x±1n ]. Then trop(X) is the support of a pure d-dimensional polyhedral

complex Σ. Let σ be a d-dimension polyhedron in Σ, and fix w in the relativeinterior of σ. The multiplicity of the polyhedron σ is the sum

mσ =∑P⊂S

mult(P, S/(inw(I))

where the sum is over all prime ideals P of S. If inw(I) is radical, then mσ is thenumber of irreducible components of V (inw(I)).

Example: When f = x2 + xy+ ty+ 1, the initial terms of f corresponding to eachone is shown in Figure 28.

Then V (y+1) is irreducible, so this cone has multiplicity one. We have V (xy+y) =V (x+1) is also irreducible, so this cone gets multiplicity one. The varieties V (xy+1)


x2 + 1

y2 + 1

y2 + xy2

x2 + xy2

Figure 29.

and V (x2 +xy) = V (x+ y) are also irreducible, but V (x2 +1) = V (x+ i)∪V (x− i),so the multiplicity of this last cone is two. This justifies Equation 2.

Definition 11.3. A weighted polyhedral complex is a polyhedral complex Σ with apositive integer on each top-dimensional polyhedron in Σ.

Let Σ be a pure d-dimensional weighted polyhedral complex, and let σ be a (d−1)-dimensional polyhedron in Σ. Let V be the subspace aff(σ)−w for w in the relativeinterior of σ. Then V lies in the lineality space of starΣ(σ), so starΣ(σ)/V is a fanin Rn/V . Let uτ ∈ Rn/V be the image of the cone τ ∈ starΣ(σ) in Rn/V for ad-dimensional polyhedron τ ∈ Σ with σ a face of τ . The polyhedral complex Σ isbalanced at σ if ∑

τ

wτuτ = 0,

where wτ is the multiplicity of the polyhedron τ .The complex Σ is balanced if it is balanced at every (d−1)-dimensional polyhedron.

Example: Let X = V (x + y + z + 1) ⊂ T 3K for K = C{{t}}. Then trop(X) is

a two-dimensional fan with rays {(1, 0, 0), (0, 1, 0), (0, 0, 1), (−1,−1,−1)} and conesspanned by any two of these. For σ the ray through (1, 0, 0) the fan startrop(X)(σ)is the one-dimensional fan in R3/ span((1, 0, 0)) with rays the images of (0, 1, 0),(0, 0, 1), and (−1,−1,−1). If we identify R3/ span((1, 0, 0)) with R2 by projectingonto the last two coordinates of R3, these are the vectors (1, 0), (0, 1), and (−1,−1),which sum to zero, so trop(X) is balanced at σ.

Theorem 11.4. Let X ⊂ T nK be an irreducible variety. Then trop(X) together withthe multiplicity is a balanced weighted polyhedral complex.

Example: Let f = x2 + y2 + xy2 + 1. Then trop(V (f)) is shown in Figure 29,together with the corresponding initial terms. Then

40 DIANE MACLAGAN

2

2

1

1

Figure 30.

V (x2 + 1) = V (x+ i) ∪ V (x− i)

V (y2 + 1) = V (y + i) ∪ V (y − i)

V (y2 + xy2) = V (x+ 1)

V (x2 + xy2) = V (x+ y2)

Thus the multiplicity on the first two cones is two, and on the second two is one.This is shown in Figure 30. Since

2

(01

)+ 2

(10

)+

(0

−1

)+

(−2−1

)=

(00

),

trop(X) is balanced.

12. Lecture 10

In this lecture we will consider a large class of examples whose tropical varietiescan be completely described. These are the X ⊂ T n whose equations are all linear.

Example 12.1. Let X = V (3x1 + 5x2 − 7x3) ⊂ T 3K where K = C{{t}}. Then

trop(X) = {(w1, w1, w2) ∈ R3 : w1 = w2 ≤ w3 or w1 = w3 ≤ w2 or w2 = w3 ≤ w1}This is the union of the three cones span((1, 1, 1)) + pos((1, 0, 0)), span((1, 1, 1)) +pos((0, 1, 0)), and span((1, 1, 1)) + pos((0, 0, 1)). Here span((1, 1, 1)) + pos((1, 0, 0))is the set {a(1, 1, 1) + b(1, 0, 0) : a, b ∈ R, b ≥ 0}. Note trop(X) is a fan, ratherthan just a polyhedral complex. This will be the case whenever the coefficients ofthe polynomials defining the ideal of X live in a subfield of K (such as C ⊂ C{{t}})where all elements have valuation zero. Note also that (1, 1, 1) lies in the linealityspace of all cones in trop(X). This will be the case whenever the equations definingthe ideal of X are all homogeneous.

We assume here that there is an inclusion i : k → K of the residue field k = R/minto our field K. This is true when K = C{{t}}. Let S = k[x±1

1 , . . . , x±1n ], and let

I = 〈f1, . . . , fr〉 ⊂ S be an ideal in S minimally generated by linear forms f1, . . . , fr.We now describe the tropical variety of X = V (I) ⊂ T .


1 2 3 4 5

123 14 15 24 25 34 35 45

Figure 31. The lattice of flats for the linear space of Example 12.2

Example 12.2. Let I = 〈x1+x2+x3+x4+x5, x1+2x2+3x3〉 ⊂ k[x±11 , x±1

2 , x±13 , x±1

4 , x±15 ].

Then V (I) is a two-dimensional subvariety of T ∼= (k∗)5.

Let fi = ai1x1 + . . . ainxn for 1 ≤ i ≤ r. Let A be the r × n matrix with entriesaij ∈ k, and let B be a (n − r) × n matrix whose rows are a basis for ker(A).Thus V (I) is equal to the intersection of the row space of B with the torus T . LetB = {b0, . . . ,bn} ⊂ kn−r be the columns of the matrix B. While B depends on thechoice of the matrix B, it is determined up to the action of GL(n− r,k).

The lattice of flats L(B) of the linear space row(B) has elements the subspaces(flats) of kn−r spanned by subsets of B. We make L(B) into a poset (partiallyordered set) by setting S1 � S2 if S1 ⊆ S2 for two subspaces S1, S2 of kn−r spannedby subsets of B. The poset L(B) is actually a lattice of rank n− r. This means thatevery maximal chain in L(B) has length n− r. See, for example, [Sta97, Chapter 3]for more on lattices.

Example 12.3. We continue Example 12.2. In this case the matrices A and B are

A =

(1 1 1 1 11 2 3 0 0

), B =

−2 1 0 1 0−2 1 0 0 1−1 −1 1 1 0

.

We thus have b1 = (−2,−2,−1), b2 = (1, 1,−1), b3 = (0, 0, 1), b4 = (1, 0, 1),and b5 = (0, 1, 0). There are fifteen subspaces of k3 spanned by subsets of B ={(−2,−2,−1), (1, 1,−1), (0, 0, 1), (1, 0, 1), (0, 1, 0)}. These are

{0} ∪ {span(bi) : 1 ≤ i ≤ 5}∪{span(b1,b2,b3), span(b1,b4), span(b1,b5), span(b2,b4),

span(b2,b5), span(b3,b4), span(b3,b5), span(b4,b5)} ∪ k3.

This gives the lattice shown in Figure 31.

A simplicial complex on a set S is a collection of subsets of S (“simplices”) thatis closed under taking subsets. For example, if S = {1, 2, 3, 4, 5}, then one simplicialcomplex is

∆ = {{1, 2, 3}, {2, 3, 4}, {3, 4, 5}, {1, 2}, {1, 3}, {2, 3}, {2, 4}, {3, 4}, {3, 5}, {4, 5},{1}, {2}, {3}, {4}, {5}, {∅}}.

42 DIANE MACLAGAN

51

4

3

2

Figure 32.

We draw simplical complexes as collections of triangles (or higher-dimensional sim-plices) in Rn. For example, the above simplicial complex is illustrated in Figure 32.

There is a simplicial complex, called the order complex, associated to any poset. Achain in a poset is a set x0 ≺ x1 ≺ · · · ≺ xs. We say that this chain has length s. Theorder complex has vertices the elements of the poset, and simplices all proper chains,which are totally ordered subsets of the poset not using the bottom or top elements(0 or Rn in our case). The order complex of L(B) is pure of dimension n− r− 2. Inthe case of our poset there is a nice geometric realization of this simplicial complex,which we now describe.

Definition 12.4. Let ei be the ith standard basis vector on Rn. Given a subsetσ ⊂ {1, . . . , n} we set eσ =

∑i∈σ ei. If V is a subspace of Rn−r spanned by some of

the bi, set σ(V ) to be {i : bi ∈ V }. Let ∆(B) be the fan whose cones are

pos(eσ(Vi) : 1 ≤ i ≤ s) + span(1),

where V1 ≺ V2 ≺ · · · ≺ Vs is a chain in L(B), and 1 is the all-ones vector in Rn.

Example 12.5. We continue Example 12.2. The fan ∆(B) ⊂ R5 has lineality spacethe span of 1 = (1, 1, 1, 1, 1), meaning that every cone contains the span of this ray, sowe describe the quotient fan in R5/1. This has 13 rays, corresponding to the five raysspanned by the bi and the eight planes spanned by them. There is a two-dimensionalcone for every inclusion of a ray into a plane, of which there are 17 in total.

The point of this construction is that the tropical variety trop(V (I)) is equal to∆(B).

Theorem 12.6. Let I be a linear ideal in S. The tropical variety of X = V (I) ⊂ Tis equal to ∆(B).

Proof. We first show that trop(X) ⊆ ∆(B). Suppose w 6∈ ∆(B). Let W j = {bi :wi ≥ j}. Note that there are only finitely many subspaces span(W j) as j varies, andthese form a chain in the lattice of flats L(B). Let l = min{j : there exists bi ∈span(W j) \W j}. If no such l existed, then w would live in the cone of ∆(B) definedby the chain {span(W j)} ⊂ L(B). Let F = span(W ll). Pick bk ∈ F \W l. Thenwk < l by the definition of W l. Since {bi : i ∈ W l} spans F , we can write bk =∑

i:bi∈W l λibi for λi ∈ k. This means that ek −∑λiei ∈ ker(B) = row(A). Thus

f = xk −∑

i∈W l λixi ∈ I. Now inw(f) = xk, so inw(I) = 〈1〉, and so w 6∈ trop(X).We next show that ∆(B) ⊆ trop(X) by exhibiting for each w ∈ ∆(B) an element

y ∈ X with val(y) = w. Given w ∈ ∆(B), let V1 ⊂ V2 ⊂ · · · ⊂ Vn−r = kn−r be the


chain of flats labelling a maximal cone of ∆(B) containing w, so dim(Vi) = i. Pickbi1 ∈ V1, and bij ∈ Vj \ Vj−1 for 2 ≤ j ≤ n− r. Note that

(3) wij ≥ wij+1.

After renumbering if necessary we may assume that ij = j, and that the matrix Bhas been chosen so that the first (n− r)× (n− r) square submatrix is the identity,which is possible as the bij are linearly independent by construction. This implies(exercise!) that the last r × r submatrix of A must be invertible, so we may assumethat it is the identity matrix (since performing row operations on A corresponds tochoosing a different generating set for I). We then have

A =(A′ Ir

), B =

(In−r −A′T

).

Set y = (tw1 , . . . , twn−r)B. Explicitly, yi = twi for 1 ≤ i ≤ n−r. For n−r+1 ≤ i ≤ n,yi =

∑n−rj=1 −aj(i−n+r)t

wj . Then Ay = 0 by construction, so y ∈ X. The valuation

val(yi) = wi for 1 ≤ i ≤ n − r by construction. For n − r + 1 ≤ i ≤ n we haveval(yi) = min{wj : a(i−n+r)j 6= 0, 1 ≤ j ≤ n − r} = ws for s = max{j : a(i−n+r)j 6=0, 1 ≤ j ≤ n− r}, by Equation 3. Now if val(yi) = wj, then bi ∈ Vj \ Vj−1, so by thechoice of w we have wi = wj, and thus val(yi) = wi. So y is the desired element ofX with val(y) = w, and so ∆(B) ⊆ trop(X). �

Remark 12.7. To visualize these tropical varieties we can do two tricks. Firstly,since we are assuming that the ideals are homogeneous, we can quotient out by thelineality space span 1 when drawing pictures. Dehomogenizing by setting one variableequal to one has the same effect. Secondly, since the tropical varieties are fans, wecan intersect the fan with the sphere Sn−1 ⊂ Rn to get an (abstract) polyhedralcomplex of dimension dim(X)− 2.

13. Lecture 11

The goal of this lecture is to describe how to compute tropical varieties in general.We begin with an example of the algorithm outlined for linear varieties last lecture.

Example: Let X = V (x0 − x1 + x3, x0 − x2 + x4, x1 − x2 + x5) ⊂ T 6. Then

A =

1 −1 0 1 0 01 0 −1 0 1 00 1 −1 0 0 1

, and B =

−1 0 0 1 1 00 −1 0 −1 0 10 0 −1 0 −1 −1

.

If we add the additional first row of (1, 1, 1, 0, 0, 0) to B, which lies in the row spaceof B, then the columns of B are the positive roots of the root lattice of A3. We have

B =

−1

00

,

0−1

0

,

00

−1

,

1−1

0

,

10

−1

,

01

−1

.

The lattice of flats L(B) is then illustrated in Figure 33. Here 013 means span(b0,b1,b3).The order complex then has 18 cones, which are indexed by the edges in Figure 33connecting two proper nontrivial flats. The cones in trop(X) ⊂ R6 are then three-dimensional, as expected. When we quotient by the lineality space span(1), we get atwo-dimensional fan in R5 ∼= R6/ span(1). Intersecting this with the sphere S4 ⊂ R5

gives a the graph that is shown in Figure 34. This is a refinement (subdividing the

44 DIANE MACLAGAN

14

0 2 3 5

013 024 05 125 23 345

∅

012345

1 4

Figure 33.

4

0

05

5

024

23 23013

125 345

1 14

Figure 34.

edges 05, 14, and 23) of a well-studied graph called the Peterson graph.We now describe how to use a computer to compute tropical varieties. That first

raises the following question:Question: What sort of examples can we expect to do on a computer?

We cannot enter a general Puiseux series into a computer, as it cannot be describedby a finite amount of information. Similarly, we cannot describe most real numbersin finite space. This suggests that the best field we can hope to work with is thealgebraic closure Q(t) of the ring of rational functions in t with coefficients in Q.Note that every example we have seen so far as lived in this field!

Any ideal with coefficients in Q(t) actually has generators with coefficients in somefinite extension L of Q(t), since the ideal is finitely generated, and a finite generating


(1,−1)

(0, 0)

Figure 35.

set has only a finite number of coefficients. We can write L ∼= Q(t)[a1, . . . , as]/J ,where J is an ideal. As a point of comparison, recall that C = R[x]/〈x2 + 1〉. We’llnow discuss how to reduce this case to computing in Q[x±1

1 , . . . , x±1n ].

If I ⊂ Q(t)[x±11 , . . . , x±1

n ], then let J = I ∩Q[t±1, x±11 , . . . , x±1

n ]. Fix w ∈ Rn. Thenw ∈ trop(V (I)) ⊂ Rn, where V (I) ⊂ T n, if and only if (1, w) ∈ trop(V (J)) ⊂ Rn+1,where V (J) ⊂ T n+1. This is the case because if p(t) is a polynomial in t, thenval(p(t)) is the smallest exponent of t occuring in p.

If I ⊂ L[x±11 , . . . , x±1

n ] where L = Q(t)[a1, . . . , as]/J is a finite extension of Q(t),then let vi = val(ai). Consider the map

φ : Q[a±11 , . . . , a±1

s , t±1, x±11 , . . . , x±1

n ] → L[x±11 , . . . , x±1

n ].

Fix w ∈ Rn. Then w ∈ trop(V (I)) if and only if (v1, . . . , vs, 1, w) ∈ trop(V (J)),where J = φ−1(I), and V (J) ⊂ T n+s+1.

Example: Let f = 1 + x + xy + ty ∈ C{{t}}[x±1, y±1] or in Q(t)[x±1, y±1]. Thentrop(V (f)) is shown in Figure 35. Let g = 1 + x + xy + ty ∈ Q[t±1, x±1, y±1]. Thentrop(g) = min(0, x, x+ y, y + 1), and

trop(V (g)) = {(w1, w2, w3) :w2 = 0 ≤ w2 + w3, w1 + w3 or

w2 + w3 = 0 ≤ w2, w1 + w3 or

w1 + w3 = 0 ≤ w2, w2 + w3 or

w2 = w2 + w3 ≤ 0, w1 + w3 or

w2 = w1 + w3 ≤ 0, w2 + w3 or

w2 + w3 = w1 + w3 ≤ 0, w2}.

For example {(w1, w2, w3) : w2 = 0 ≤ w2 + w3, w1 + w3} = {(w1, w2, w3) : w2 =0, w3 ≥ 0, w1 + w3 ≥ 0} = pos((−1, 0, 1), (1, 0, 0)).

46 DIANE MACLAGAN

B

D

A

(0, 0)

C

(1,−1)

E

Figure 36.

Then

trop(V (g)) ∩ {w1 = 1} ={(1, 0, w3) : w3 ≥ 0}∪{(1, w2,−w2) : 0 ≤ w2 ≤ 1}∪{(1, w2,−1) : w2 ≥ 1}∪{(1, w2, 0) : w2 ≤ 0}∪∅∪{(1, 1, w3) : w3 + 1 ≤ 0}.

Labelling these cones A through E, these are shown in Figure 36.These reductions mean that if I ⊂ Q(t)[x±1

1 , . . . , x±1n ] then we can compute trop(V (I))

by doing computations in a Laurent polynomial ring with coefficients in Q, so weneed only discuss how to compute tropical varieties in this context.

A first algorithm to compute trop(V (I)) in this context is to first homogenize theideal I to get an ideal Ih ∈ Q[x0, . . . , xn]. We can then compute the Grobner fanof Ih, and throw away those cones containing some w in their relative interior withinw(Ih) containing a monomial. A problem with this, however, is that it is quiteinefficient, as there can be many more cones in the Grobner fan than there are inthe tropical variety. In [BJS+07] an example is given of a family of ideals in threevariables indexed by p ∈ N where the tropical varieties all have four rays, but theGrobner fan of the ideal Ip has at least 1/4(p+ 1) rays.

The key fact for the algorithm used by gfan is the following theorem.

Theorem 13.1. [BJS+07] Let X ⊂ T nK, where char(K) = 0, be an irreducible varietyof dimension d. Then X is a pure polyhedral complex of dimension d that is connectedin codimension one.

Here “connected in codimension one” means that there is a path

σ = σ1 � τ1 ≺ σ2 � τ2 ≺ σ3 � τ3 . . . τs ≺ σs = σ′

connecting any two d-dimensional polyhedra σ, σ′ of trop(X), where τ ≺ σ means “τis a facet ((d-1)-dimensional face) of σ. This is illustrated in Figure 37.


Connected in codimension one Not connected in codimension one

Figure 37.

The algorithm used by gfan starts by computing a d-dimensional cone in theGrobner fan of Ih for which the initial ideal does not contain a monomial. For eachfacet of this cone, it computes the neighbouring cones in the tropical variety, andthen “walks” around the tropical variety until all cones have been visited. Theo-rem 13.1 guarantees that every cone will be found in this fashion. The computationof computing neighbouring cones can be effectively reduced to the case d = 1. Formore details, see [BJS+07].

The program gfan [Jen], written by Anders Jensen, can be accessed via yourschwartz accounts, or by using Andrew Hoefel’s applet available at

http://www.mathstat.dal.ca/~handrew/gfan/index.php

To use it, create a file containing content like:

Q[x_1,x_2,x_3,x_4,x_5]

{x_1+x_2+x_3+x_4+x_5, x_1+2*x_2+3*x_3}

Note that unlike M2 there is only one Q in the name of the rational numbers. Toget the starting cone, type gfan tropicalstartingcone < myfile >myfile.cone,where myfile is the name of the file you created above, and you can change the nameof myfile.cone. This will give output like:

Q[x_1,x_2,x_3,x_4,x_5]

{

x_2+2*x_3,

x_1-x_3}

{

x_2-x_5-x_4+2*x_3,

x_1+2*x_5+2*x_4-x_3}

This is a list of the initial ideal, and then the ideal. If you are using the web-applet,you will enter the input in the dialogue box, and the output will be on the screenbelow, which you will need to paste back into the dialogue box for the next step.Next type gfan tropicaltraverse <myfile.cone >myfile.output. This will giveoutput of the form:

_application PolyhedralFan

_version 2.2

_type PolyhedralFan

AMBIENT_DIM

5

48 DIANE MACLAGAN

DIM

3

LINEALITY_DIM

1

RAYS

-1 0 0 0 0 # 0

0 -1 0 0 0 # 1

-1 -1 -1 0 0 # 2

0 0 -1 0 0 # 3

1 1 1 1 0 # 4

0 0 0 -1 0 # 5

N_RAYS

6

LINEALITY_SPACE

1 1 1 1 1

ORTH_LINEALITY_SPACE

0 0 0 1 -1

0 0 1 0 -1

0 1 0 0 -1

1 0 0 0 -1

F_VECTOR

1 6 10

CONES

{} # Dimension 1

{0} # Dimension 2

{1}

{2}

{3}

{4}

{5}

{0 2} # Dimension 3

{2 3}

{1 2}

{0 4}

{1 4}

{0 5}

{1 5}

{4 5}


{3 4}

{3 5}

MAXIMAL_CONES

{0 2} # Dimension 3

{2 3}

{1 2}

{0 4}

{1 4}

{0 5}

{1 5}

{4 5}

{3 4}

{3 5}

PURE

1

MULTIPLICITIES

1 # Dimension 3

1

1

1

1

1

1

1

1

1

This says that the tropical variety has lineality space span(1), and is three-dimensionalThe quotient by the lineality space gives a fan with six rays, and ten two-dimensionalcones, which are listed.

Note that this is not what we got when we did this example on Friday. Firstly thisis because gfan uses the max convention instead of the min convention. Secondly, itis because gfan has amalgamated some of the cones we constructed. We had rays eifor 1 ≤ i ≤ 5, e1 + e2 + e3, and then ei + ej for 1 ≤ i < j ≤ 5 with {i, j} 6⊂ {1, 2, 3}.Two of the cones were span(1)+pos(e1, e1 +e4), and span(1)+pos(e4, e1 +e4). Theunion of these two cones is span(1) + pos(e1, e4).Warning: The program gfan assumes that the ideal generated by your inputpolynomials is prime, so the variety is irreducible. It will not (necessarily) give thecorrect answer if your input variety is not irreducible.

Tropical Bases.

50 DIANE MACLAGAN

Definition 13.2. Let I ⊆ K[x±11 , . . . , x±1

n ]. A set {f1, . . . , fs} ⊂ I is a tropical basisfor I if

trop(V (I)) =s⋂i=1

trop(V (fi)).

Example: Let I = 〈x + y + z, x + 2y〉 ⊂ C{{t}}[x±1, y±1, z±1]. Then trop(V (I)) =span(1). Any two polynomials in the set

{x+ 2y, y − z, x+ 2z}form a tropical basis for I, as trop(V (x+2y)) = {w ∈ R3 : w1 = w2}, trop(V (y−z)) ={w ∈ R3 : w2 = w3}, and trop(V (x + 2z)) = {w ∈ R3 : w1 = w3}. The set {x + y +z, x+2y} is not a tropical basis, as the vector (0, 0, 1) ∈ trop(V (x+y+z))∩trop(V (x+2y)), but (0, 0, 1) 6∈ trop(V (I)). Note that trop(V (x + y + z)) ∩ trop(V (x + 2y)) isnot a balanced polyhedral complex.

Example: Let I = 〈f〉 be a principal ideal. Then {f} is a tropical basis for I.This was Q4 of Exercises 3.

Proposition 13.3. Let I ⊆ K[x±11 , . . . , x±1

n ]. Then I has a finite tropical basis.

Proof. Let Ih be the homogenization of I in K[x0, . . . , xn], and let Σ be the Grobnercomplex of Ih. There are only finitely many polyhedra σ ∈ Σ, and for each σ ∈ Σthe initial ideal inw(Ih) is constant for all w in the relative interior of σ, so we maydenote it by inσ(I

h). Given σ ∈ Σ with inw(Ih) containing a monomial for w in the

relative interior of σ (so inw(I) = 〈1〉), we can find fσ ∈ Ih with inw(fσ) a monomial.

Let fσ ∈ K[x±11 , . . . , x±1

n ] denote the dehomogenization of fσ, and let

G = {fσ : inσ(Ih) contains a monomial }.

Note that G is a finite set, and⋂fσ∈G

trop(V (fσ)) ⊆ trop(V (I)),

since relint(σ)∩trop(V (fσ)) = ∅. But since fσ ∈ I, we have trop(V (I)) ⊆ trop(V (fσ)),so ⋂

fσ∈G

trop(V (fσ)) = trop(V (I)).

�

Open Problem: Give a good algorithm to compute a tropical basis.There is an algorithm implicit in the above proof, but it involves knowing the

Grobner complex of Ih, which is hard to compute.

Definition 13.4. Suppose k includes into K. Let I ⊂ k[x±11 , . . . , x±1

n ] be a linearideal (generated by linear polynomials). A linear polynomial f =

∑ni=1 aixi is a

circuit for I if {i : ai 6= 0} is minimal with respect to inclusion. This means thatthere is no g =

∑nj=1 bjxj ∈ I with {j : bj 6= 0} ( {i : ai 6= 0}. Note that there are

only a finite number of circuits up to scaling.

Exercise: Let I be a linear ideal. Then the set of circuits form a tropical basis forI.


Remark 13.5. There is also a stricter notion of tropical basis, where we requirethat the tropical basis in addition be a Grobner basis for I with respect to anyw ∈ trop(V (I)). This is also finite.

14. Lecture 12

In this lecture we discuss the tropicalization of the Grassmannian. We first reviewthis beautiful classical variety.

Let V be an n-dimensional vector space. We will use k to denote the field of V bydefault, but the constructions will make sense over an arbitrary vector space. TheGrassmannian G(d, n) parameterizes all d-dimensional subspaces of V ∼= kn. It is asmooth projective variety of dimension d(n− d) (so “locally” looks like affine spaceof dimension d(n− d)), whose points correspond to d-dimensional subspaces of V .Example: The Grassmannian G(1, n) parameterizes all one-dimensional subspacesof V , which are lines through the origin in kn. Thus G(1, n) ∼= Pn−1.

We’ll describe two ways to see that G(d, n) is a variety.Approach 1: We’ll describe G(d, n) by giving an affine cover. An affine coveris open cover (in the Zariski topology) of the set G(d, n) by affine varieties. It isanalogous to giving charts for a manifold.

Fix a basis for V , and thus an isomorphism of V with kn. Given a d-dimensionalsubspace W ⊆ V ∼= kn, choose a basis for W and write this basis, in the basis for V ,as the rows of a d× n matrix Wmat with entries in k. For each set σ ⊂ {1, . . . , n} ofsize d we consider the set

Gσ = {W ⊂ V : the submatrix of Wmat indexed by the columns of σ is invertible}.Given W ∈ Gσ, there is a unique basis for W for which Wmat|σ is the identity matrix.Example: Consider G(2, 4), and σ = {1, 2}. Then if W ∈ Gσ, then there is achoice of basis for W in which

(4) Wmat =

(1 0 a b0 1 c d

),

where a, b, c, d ∈ k. So Gσ∼= A4. We can glue together all the different Gσ to get a

variety.Example: G12 ∩G13. Then if W ∈ G12 ∩G13 then there is a choice of basis for Was above, where c 6= 0. Thus G12 ∩ G13

∼= A3 × k∗. There is also a version of Wmat

of the form

Wmat =

(1 −a/c 0 b− ad/c0 1/c 1 d/c

).

We identify the two copies of A4 on their overlap using the above transitions.

Approach 2: We can embed G(d, n) into P(nd)−1 by sending W ∈ G(d, n) to the

vector of d× d minors of Wmat. This is the Plucker embedding of G(d, n). Note thatif we choose a different basis for W , then the vector of minors changes by at most a

multiplicative constant, so the map to P(nd)−1 is well-defined. This map is injective,

and the image is a subvariety cut out by the Plucker relations, which we explain inthe case d = 2.Example: Consider G(2, 4). Then

(42

)= 6. Label the coordinates of P5 as

x12, x13, x14, x23, x24, x34. The Plucker relation is x12x34−x13x24+x14x23. For example,

52 DIANE MACLAGAN

for the matrix Wmat of Equation 4, the Plucker image is (1 : c : d : −a : −b : ad− bc).This satisfies the Plucker relation.

Theorem 14.1. Let

I2,n = 〈pijkl = xijxkl−xikxjl+xilxjk : 1 ≤ i < j < k < l ≤ n〉 ⊂ k[xij : 1 ≤ i < j ≤ n].

Then G(2, n) = V (I2,n) ⊂ P(n2)−1.

There is a similar description for G(d, n) for d > n.

Let AG(2, n) = V (I2,n) ⊂ A(n2). Then AG(2, n) is the affine cone over the Grass-

mannian G(2, n). Let T be torus of A(n2). Then T = {(x12, x13, . . . , x(n−1)n) ∈ A(n

2) :xij 6= 0}. Let X = AG(2, n) ∩ T .Claim: The set of Plucker relations forms a tropical basis for I2,n ⊆ K[x±1

ij ], so

trop(X) =⋂

1≤i<j<k<l≤n

trop(V (pijkl)).

For a proof, see [SS04].

Thus trop(X) = {w ∈ R(n2) : for all 1 ≤ i < j < k < l ≤ m either wij + wkl =

wik+wjl ≤ wil+wjk, or wij +wkl = wil+wjk ≤ wik+wjl, or wik+wjl = wil+wjk ≤wij + wkl}. This is the space of Phylogenetic trees.

Definition 14.2. A tree is a graph with no cycles. The degree of a vertex in a treeis the number of edges incident to that vertex. A tree is trivalent if every vertexhas degree three or one. The vertices of degree one are called leaves. A tree is leaf-labelled if all leaves have a label from some set S. A phylogenetic tree is a trivalentleaf-labelled tree.

The name comes from the fact that diagrams illustrating closeness of species aretrivalent graphs. See Figure 38.

Given a phylogenetic tree τ drawn in the plane, we can record the tree-distancebetween two vertices.Example: For the tree of Figure 39, the tree-distance between vertices 1 and 2 isa+ b. Ordering the pairs (i, j) as 12, 13, 14, 23, 24, 34, we have the distance betweenvertices i and j recorded in the vector

(a+ b, a+ c+ d, a+ c+ e, b+ c+ d, b+ c+ e, d+ e).

This equals

a

111000

+ b

100110

+ c

011110

+ d

010101

+ e

001011

.


Primordial life

1 3

2 4

1

Sparrow Gorilla Human

6

23

4

5

Figure 38. Some phylogenetic trees

c

a

b

d

e

1

2

3

4

Figure 39.

If we let the leaf lengths become negative, this cone is then

span

111000

,

100110

,

010101

,

001011

+ pos

011110

,

which equals

{w : w13 + w24 = w14 + w23 ≥ w12 + w34}.In general, a vector is the tree-distance vector for a trivalent tree with n leaves

if and only if it satisfies the four point condition: for any four leaves i, j, k, l, ifwe form the three possible sums of distances between disjoint pairs d(i, j) + d(k, l),

54 DIANE MACLAGAN

3

1

3

2

4

1

4

2

Figure 40.

5

3

4

5

1

2

35

45

132534

24 14

15 23

12

1

2

3

4

5

1

23

4

Figure 41.

d(i, k) + d(j, l), and d(i, l) + d(j, k), then two of these sums are equal and greaterthan or equal to the third.

For trees with four leaves the other two options are shown in Figure 40. Setting

V = span

111000

,

100110

,

010101

,

001011

,

these cones are V + pos((1, 0, 1, 1, 0, 1)) and V + pos((1, 1, 0, 0, 1, 1)).Note that the union of these three cones is − trop(X)!

Summary: The set − trop(X) is the set of tree-distance vectors for phylogenetictrees with n leaves. This is the space of phylogenetic trees, and has one cone for eachcombinatorial type of labelled tree. It is a polyhedral fan, with an n-dimensionallineality space. The quotient by the lineality space is a pure fan of dimension n− 3.Note that n− 3 is the number of internal (not adjacent to a leaf) edges of a trivalenttree with n leaves.

When n = 4 we get the three cones described above. When n = 5 the quotient bythe five-dimensional lineality space is a two-dimensional fan in R5. The intersectionof this with the sphere S4 is the graph shown in Figure 41.

For more details on all this, see [SS04].


15. Exercises

These questions cover Wednesday of week two to Tuesday of week three.

(1) Show that the definition of the star of a polyhedron σ in a polyhedral complexΣ does not depend on the choice of the point w ∈ σ.

(2) Let A be an r×n matrix of rank r with entries in k, and let B be a (n−r)×nmatrix whose rows form a basis for ker(A). Show that if the first (n − r) ×(n − r) submatrix of B is invertible, then the last r × r submatrix is alsoinvertible.

(3) For each of the following polynomials f(a) Compute trop(V (f)).(b) For each polyhedron σ in the polyhedral complex trop(V (f)) compute

inw(I) for any w in the relative interior of σ, and compute directlytrop(V (inw(I)). Check that this is the appropriate star.

(c) Check that trop(V (f)) is a weighted balanced polyhedral cone. (Thepolynomials have been chosen so that all weights are one).

(I do not expect to see all of these written up completely - do enough untilyou are satisfied you understand what is going on).(a) f = x+ ty + t3 ∈ C{{t}}[x±1, y±1];(b) f = x2 + txy + x+ y + t ∈ C{{t}}[x±1, y±1];(c) f = x+ y + x2y + xy2 + x2y2 ∈ C{{t}}[x±1, y±1];(d) f = x+ y + z + 1 ∈ C{{t}}[x±1, y±1, z±1].

(4) For each of the following linear varieties compute trop(X). Repeat the veri-fications of parts (b) and (c) of the previous question in this context.(a) X = V (x1 + x2 + x3 + x4, x1 + 2x2 + 4x3 − x4) ⊂ T 4;(b) X = V (x1 + x2 + x3 + x4 + x5, x1 − x2 + 3x3 + 4x4 + 7x5) ⊂ T 5;(c) X = V (x1 + x2 + x3 + x4 + x5, x1 + x2 + x3 + 3x4 − x5) ⊂ T 5.

(5) Let X ⊂ T n be defined by linear equations with coefficients in k.. Show thatthe multiplicity of each cone in trop(X) is one.

(6) The goal of this exercise is to explain in detail how to compute inw(I) usinga computer algebra package such as Macaulay 2.(a) Read the proof of Lemma 8 and Corollary 9 of Lecture 8.

(b) Given I = 〈f1, . . . , fr〉 ⊆ k[x±11 , . . . , x±1

n ]. Let fi be the polynomial ob-

tained from fi by clearing denominators. Check that I = 〈f1, . . . , fr〉 ⊂k[x±1

1 , . . . , x±1n ].

(c) Let I = (〈f1, . . . , fr〉 : (∏n

i=1 xi)∞) ⊂ k[x1, . . . , xn]. Show that I =

I ∩ k[x1, . . . , xn].(d) Let I = 〈x2−y, y2−x〉 ⊂ K[x±1, y±1]. Use I to show that the saturation

step is necessary in the previous exercise.(e) Let f ′i be the homogenization of fi using the variable x0. Let J =

(〈f ′1, . . . , f ′r〉 : x∞0 ) ⊆ k[x0, . . . , xn]. Show that V (J) ⊂ Pn is the clo-sure of the image under the map x 7→ (1 : x) of V (I) ⊂ An. (Thesaturation step is again necessary here: an optional exercise is to find anexample showing this).

(f) Show that for most choices of v ∈ Rn we have that inv(inw(J)) is amonomial ideal.

56 DIANE MACLAGAN

(g) Use the previous parts of this exercise to describe how to use a computeralgebra package to compute inw(I).

(h) In Macaulay 2 we describe a ring usingR=QQ[x_0..x_5, Weights=>{1,2,3,4,5,6}].

In this example w = (1, 2, 3, 4, 5, 6). If f =∑aux

u, let inw(f) =∑w·u is maximal aux

u. (Note the difference between min and max here!)Show that inw(f) = inw+λ1(f), where 1 ∈ Rn+1 is the all-ones vector,and λ ∈ R.The command leadTerm(1,J) computes generators for the ideal 〈inw(f) :f ∈ J〉. Weights must be positive in Macaulay 2. Conclude that ifw ∈ Rn, setting Weights=>w2 where w2 = N1 − (0, w) for N � 0 letsleadTerm(1,J) compute inw(I).

(7) The goal of this exercise is to prove that trop(X) is a weighted balancedcomplex when X is a curve in T 2. Since the balanced condition is a localcondition, we only need to consider ideals in k[x1, . . . , xn], so trop(X) is aone-dimensional fan. Let

f =∑

(i,j)∈Z2

aijxiyj,

Let

P = conv((i, j) : aij 6= 0) ⊂ R2.

where aij ∈ k. Then inner normal to an edge e of P is a vector w ∈ R2 withw · (i, j) < w · (i′, j′) if (i, j) ∈ e and (i′, j′) ∈ P \ e.(a) Let f = x+ y+x2y+xy2 +x2y2. Draw P , and the inner normal to each

edge of P . Check that this picture is trop(V (f)).Note that we can always choose the inner normal w to an edge e to be in

Z2. A vector w ∈ Z2 is primitive if gcd(w1, w2) = 1.(b) Show that every edge e ∈ P has a unique primitive inner normal we ∈ Z2.(c) Show that for any f ∈ k[x±1, y±1] the tropical variety trop(V (f)) is equal

to ∪e pos(we), where the union is over all edges e of P .(d) Let we be the primitive inner normal to an edge we of P . Show that

there is a monomial m = xayb and a polynomial g ∈ k[m] for which

V (inwe(f) = V (g)

and deg(g) is the lattice length of the edge e (one less than the numberof lattice points in e). Thus conclude that the multiplicity me of the raypos(we) of trop(V (f)) is the lattice length of e.

(e) Verify the previous question for f = x+ y + x2y + xy2 + x2y2.(f) Conclude that

∑emewe = 0, so trop(V (f)) is a balanced weighted poly-

hedral fan. Hint: Let e be the vector corresponding to the edge e, ori-ented clockwise. Note that

∑e e = 0.

(8) Let f = tx2 + xy + ty2 + x + y + t ∈ C{{t}}[x±1, y±1]. Compute trop(V (f)),and compare with trop(V (g))∩ {w1 = 1} for g = tx2 + xy+ ty2 + x+ y+ t ∈C[t±1, x±1, y±1].


(0, 0)

Figure 42.

(9) Play with gfan. Plug in some of the examples we have computed in lecturesand from the notes and check that gfan agrees. Warning: Remember thatgfan uses the max convention instead of our min convention.

(10) Use gfan to compute the tropical variety of X = V (I) ⊂ T 10, where I =〈x12x34−x13x24+x14x23, x12x35−x13x25+x15x23, x12x45−x14x35+x15x24, x13x45−x14x35+x15x34, x23x45−x35x24+x25x34〉 ⊂ k[x±1

12 , x±113 , x

±114 , x

±115 , x

±123 , x

±124 , x

±125 , x

±134 , x

±135 , x

±145 ].

Also compute the tropical variety of X = V (J) ⊂ T 6 where J = 〈1 − y1 +y3, 1−y2+y4, y1−y2+y5〉 ⊂ k[y±1

1 , . . . , y±15 ] (Hint: you will need to homogenize

first). Compare your answers. Can you explain what you observe?(11) Let X = V (xy+y2 +2xz+2yz−xw−yw, x2−y2−3xz−3yz+3xw+3yw) ⊂

T 4. Show that X is not irreducible. Hint: The decompose and intersect

commands in M2 may help. What does gfan think is the tropical variety ofX? Is this correct?

(12) (Only for the very computationally inclined). Look at the paper math.AG/0507563for details of the algorithm gfan uses to compute tropical varieties.

(13) Describe the tropical variety of the Grassmannian G(2, 6). How many conesof each dimension are there?

16. Lectures 13, 14, and 15

For the rest of the week we consider the closure of X ⊂ T n in a toric variety. Westart with taking the closure of X in An. We will consider only the case that k → K,such as C → C{{t}}.

Recall that T n ⊂ An, since T n = {(x1, . . . , xn) : xi ∈ k, xi 6= 0}, and An ={(x1, . . . , xn) : xi ∈ k}. Given X ⊂ T n, let X be the closure of X in An. This is thesmallest closed set in An containing X, so X = V (I) for some I and is as small as

possible. Recall that if V (I) ( V (J) ⊂ T n, then√J (

√I. If U is a subset of An

then U =⋂U⊆V (I):I⊆k[x1,...,xn] V (I).

Question: Given X ⊂ T n is 0 ∈ X?The answer is given by the following theorem, which was first observed by Tevelev [Tev07].

Theorem 16.1. Let X ⊂ T nk , and let X be the closure of X in An. Then 0 ∈ X ifand only if trop(X) ∩ Rn

>0 6= ∅, where Rn>0 = {(x1, . . . , xn) : xi > 0 for 1 ≤ i ≤ n}.

Example: Let X = V (x+y+1) ⊂ T 2. Then X = {(a,−1−a) : a ∈ k∗, a 6= −1}, soX = {(a,−1−a) : a ∈ k} = X∪{(0,−1), (−1, 0)} = V (x+y+1) ⊂ A2. The tropicalvariety trop(X) is shown in Figure 42. Note that 0 6∈ X, and trop(X) ∩ R2

>0 = ∅.

58 DIANE MACLAGAN

Example: Let X = V (x2 − y) ⊂ T 2. Then X = {(a, a2) : a ∈ k∗}, and X ={(a, a2) : a ∈ K} = X ∪ {(0, 0)} = V (x2 − y) ⊂ A2. Then trop(X) is the linew2 = 2w1, which contains the point (1, 2) ∈ R2

>0, and 0 ∈ X.

Example: Let X = V (xy−1) = {(a, 1/a) : a ∈ k∗} ⊂ T 2. Then X = X, so 0 6∈ X.The tropical variety is the line w1 + w2 = 0, which does not intersect the positiveorthant.

We first note the following description of the ideal of X ⊂ An.

Lemma 16.2. If X = V (I) ⊂ T n for I ∈ k[x±11 , . . . , x±1

n ] then X = V (I) ⊂ An,where I = I ∩ k[x1, . . . , xn].

Proof. Let X = V (J) for J ⊂ k[x1, . . . , xn]. Then since X ⊂ X, we have f(x) = 0 for

all x ∈ X and f ∈ J , so f ∈√I when f is regarded as an element of k[x±1

1 , . . . , x±1n ].

Let J ′ = Jk[x±11 , . . . , x±1

n ]. Then J ′ ⊆√I. Now J ⊆ J ′ ∩ k[x1, . . . , xn] ⊆

√I ∩

k[x1, . . . , xn]. We claim that√I ∩ k[x1, . . . , xn] =

√I. To see this, note that if f ∈√

I ∩k[x1, . . . , xn], then there is N > 0 for which fN ∈ I, and since f ∈ k[x1, . . . , xn]

we have fN ∈ I, so f ∈√I. Conversely, if f ∈

√I, then there is N > 0 for which

fN ∈ I, so fN ∈ I, and f ∈ k[x1, . . . , xn], so f ∈√I ∩ k[x1, . . . , xn]. Thus J ⊆

√I,

so V (I) ⊆ V (J) = X. But if x ∈ X then f(x) = 0 for all f ∈ I, so X ⊆ V (I), andso X ⊆ V (I). Thus X = V (I). �

A key idea of the proof of Theorem 16.1 is the inclusion of tori. This will let usreduce to the case where X is a curve in T 2.

Definition 16.3. A morphism φ : T n → Tm is determined by a k-algebra homo-morphism φ∗ : k[y±1

1 , . . . , y±1m ] → k[x±1

1 , . . . , x±1n ]. Note that the map φ∗ goes in the

reverse direction!

A point a = (a1, . . . , an) ∈ T n corresponds to the maximal ideal

Ia = 〈x1 − a1, . . . , xn − an〉 ∈ k[x±11 , . . . , x±1

n ].

Since the induced map k[y±11 , . . . , y±1

m ] → k[x±11 , . . . , x±1

n ]/Ia ∼= k is surjective, thekernel φ∗−1(Ia) is maximal, so is of the form 〈y1 − b1, . . . , ym − bm〉 ∈ k[y±1

1 , . . . , y±1m ]

for some b = (b1, . . . , bm) ∈ Tm. We thus set φ(a) = b.Note that φ∗(yi) is an invertible polynomial in k[x±1

1 , . . . , x±1n ] for 1 ≤ i ≤ m, so

must be a monomial. We thus have

φ∗(yi) = xui for ui ∈ Zn.

Thus a morphism φ : T n → Tm corresponds to a map ψ : Zm → Zn given byψ(ei) = ui, where ei is the ith standard basis vector of Rm.Exercise: The morphism φ is surjective if and only if ψ is injective. The morphismφ is injective if and only if ψ is surjective.

We record ψ by the n × m matrix U with columns u1, . . . ,um, so ψ(v) = Uv.The map φ : T n → Tm is given by φ(t1, . . . , tn) = (tu1 , . . . , tum), where tui =tui11 tui2

2 . . . tuinn .

Given X ⊂ T n with X = V (I), the closure φ(X) of the image of X under φ is thevariety V (φ∗−1(I)).


Proposition 16.4. Let φ : T n → Tm be a morphism of tori, with associated n×mmatrix U . Let X ⊂ T n be a variety, and let φ(X) be the closure of its image in Tm.

Then w ∈ trop(X) if and only if UTw ∈ trop(φ(X)).

Proof. We denote by X(K) the subvariety of T nK defined by the same equationsas X. If w ∈ trop(X) then by the Fundamental Theorem there is y ∈ X(K) with

val(y) = w. Then φ(y) = (yu1 , . . . , yum) ∈ φ(X), and val(φ(y)) = (u1 ·val(y), . . . ,um ·val(y)) = UTw. So w ∈ trop(X) implies UTw ∈ trop(φ(X)).

Conversely, if w ∈ trop(φ(X)), then there exists y ∈ φ(X) with val(y) = w. By

[Pay07] the set of y ∈ φ(X) with val(y) = w is Zariski dense in φ(X), so we mayassume that y ∈ φ(X). Thus there is y ∈ X with y = φ(y). Then w = val(y) =val(φ(y)) = UT val(y), so there is w = val(y) ∈ trop(X) with w = UTw. �

We note that one direction of Proposition 16.4 can also be seen by arguments withinitial ideals.

Lemma 16.5. Let I = k[x±11 , . . . , x±1

n ], and let w ∈ Rn. Let φ : T n → Tm be givenby φ(t)i = tui, and let U be the n×m matrix with columns the vectors ui. Let φ∗ bethe k-algebra homomorphism k[y±1

1 , . . . , y±1m ] → k[x±1

1 , . . . , x±1n ]. Then

inUTw(φ∗−1(I)) ⊆ φ∗−1(inw(I)).

Thus w ∈ trop(X) implies that UTw ∈ trop(φ(X)).

Proof. Let f =∑

v∈Nn avyv ∈ φ∗−1(I), so φ∗(f) =

∑v∈Nn avx

Uv ∈ I. Then inUTw =∑UTw·v=W avy

v, where W = min{UTw · v : av 6= 0} = min{vTUTw : av 6= 0}. How-ever inw(φ(f)) =

∑w·Uv=W ′ avx

Uv where W ′ = min{w ·Uv : av 6= 0} = min{vTUTw :av 6= 0} = W . Thus {v : w · Uv = W, av 6= 0} = {v : UTw · v, av 6= 0}, soinw(φ(f)) = φ(inUTw(f)). Thus inUTw(f) ∈ φ−1(inw(I)).

If w ∈ trop(X), then inw(I) 6= 〈1〉, so inUTw(φ∗−1(I)) 6= 〈1〉, and thus UTw ∈trop(φ(X)). �

We next outline how to reduce the proof of Theorem 16.1 to the case where X isa curve.

Lemma 16.6. Let X ⊂ T n with ideal I ⊂ k[x±11 , . . . , x±1

n ]. Let φ : Tm → T n bea morphism of tori with associated m × n matrix U and k-algebra homomorphismφ∗ : k[x±1

1 , . . . , x±1n ] → k[y±1

1 , . . . , y±1m ]. Then Y = φ−1(X) = V (φ∗(I)) ⊆ Tm.

We have the containment of sets UT trop(Y ) ⊂ trop(X) ∩ trop(im(φ)).

Proof. Note that if f =∑

v∈Zn avxv, then φ∗(f) =

∑v∈Zn avx

Uv, so φ∗(f)(y) =f(φ(y)) for all y ∈ Tm. Let y ∈ Y . Then φ∗(f)(y) = f(φ(y)) = 0 for all f ∈ I, soy ∈ V (φ∗(I)). Conversely, if g(y) = 0 for all g ∈ φ∗(I), then φ∗(f)(y) = 0 for allf ∈ I, so f(φ(y)) = 0 for all f ∈ I, and thus φ(y) ∈ V (I) = X, so y ∈ Y .

Let w ∈ trop(Y ). Then there is y ∈ Y with val(y) = w. Thus φ(y) ∈ X,and so val(φ(y)) = UTw ∈ trop(X). Since trop(im(φ)) = imUT , it follows thatUTw ∈ trop(X) ∩ trop(im(φ)). �

The morphism φ : T n → Tm extends to a morphism φ : An → Am if and only ifevery entry of U is nonnegative. To reduce to the case thatX is a curve, one intersects

60 DIANE MACLAGAN

with a codimension dim(X) − 1 subtorus of T n that intersects X transversely in acurve Y ⊂ Tm for m = n− dim(X) + 1. We assume that the inclusion φ : Tm → Tm

has matrix U with positive entries. It then follows from Lemma 16.6 that if 0 ∈ Xthen 0 ∈ Y , so given the curve case of Theorem 16.1 we know that there is w ∈trop(Y ) with w ∈ Rm

>0. Then UTw ∈ trop(X) ∩ Rm>0.

We will only prove Theorem 16.1 in the case where X is a curve in T 2. Theelementary approach to a proof proposed in class does not work.

Proposition 16.7. Let X ⊂ T 2k be a curve, so X = V (f) for some f ∈ k[x±1, y±1].

Then 0 ∈ X if and only if trop(X) ∩ R2>0 6= ∅.

Proof. Write f =∑

(i,j)∈Z2 aijxiyj. Let I = 〈f〉 ⊆ k[x±1, y±1], and let I = I ∩ k[x, y].

Then I = 〈f ′〉 for f ′ = x−ky−lf , where k = min{i : aij 6= 0}, and l = min{j : aij 6=0}.

Now 0 ∈ X if and only if f ′(0, 0) = 0, which occurs if and only if akl 6= 0. Ifakl 6= 0, then for all w ∈ R2

>0, we have inw(f ′) = akl, so inw(I) = 〈1〉, and thustrop(X)∩R2

>0 = ∅. Conversely, if akl = 0, then let P = conv((i− k, j − l) : aij 6= 0).By construction P lives in the positive orthant, and has a vertex v1 of the form (0, r)and a vertex v∞ of the form (s, 0). Let v2 be the next vertex in counter-clockwiseorder from v1. Let w = (w1, w2) be the inner facet normal of the edge joining v1 andv2 (so w ·u ≥ w · v1 = w · v2 for all u ∈ P ). Note that w1, w2 > 0. Then inw(f ′) is nota monomial, since its support contains the monomials with exponents giving rise tov1 and v2, and so inw(f) is not a monomial, and thus w ∈ trop(X). �

Theorem 16.1 generalizes to the following theorem.

Theorem 16.8. Let X ⊂ T n and let X be the closure of X in An. Then forσ ∈ {1, . . . , n}

X ∩ {x ∈ An : xi = 0 for all i ∈ σ, xi 6= 0 for all i 6∈ σ} 6= ∅

if and only if there is w ∈ trop(X) with wi > 0 for all i ∈ σ, and wi = 0 for all i 6∈ σ.

The case σ = {1, . . . , n} is Theorem 16.1. The condition on w ∈ trop(X) can berephrased as asking that w lies in the relative interior of pos(ei : i ∈ σ). Examplesof Theorem 16.8 can be seen by considering the subvarieties of T 2 at the start of thelecture.

One can also ask the same question about the closure of X ⊂ T n in Pn. Recallthat T n embeds into Pn by the map (x1, . . . , xn) 7→ (1 : x1 : · · · : xn). Given X ⊂ T n,let X now denote the closure of X in Pn. This is the smallest projective varietycontaining X.Example: Let X = V (x1 + x2 + 1) ⊂ T 2. Let X be the closure of X in P2 underthe embedding T 2 → P2 given by (t1, t2) 7→ (1 : t1 : t2). Then X = X ∪ {(1 : 0 :−1), (1 : −1 : 0), (0 : 1 : −1)} = V (x1 + x2 + x0). Note that X ∩ {xi = 0} 6= ∅ fori = 0, 1, 2, while X ∩ {xi = xj = 0} = ∅ for all choices of 0 ≤ i < j ≤ 2.

Note that the torus T n is a group, with multiplication coordinatewise, and identitythe element (1, 1, . . . , 1) ∈ T n. The torus T n acts on Pn by

(t1, . . . , tn) · (x0 : x1 : · · · : xn) = (x0 : t1x1 : t2x2 : · · · : tnxn).


The orbits of T n on Pn are indexed by proper subsets of {0, 1, . . . , n} indicating whichcoordinates are zero.Example: The torus T 2 acts on P2 by (t1, t2) · (x0 : x1 : x2) = (x0 : t1x1 : t2x2).The orbits of T 2 on P2 are:

{T 2, {(0 : 1 : t2) : t2 6= 0}, {(1 : 0 : t2) : t2 6= 0}, {(1 : t1 : 0) : t1 6= 0},{(1 : 0 : 0)}, {(0 : 1 : 0)}, {(0 : 0 : 1)}.

These can be labelled by following subsets of {0, 1, 2}:{∅, {0}, {1}, {2}, {1, 2}, {0, 2}, {0, 1}}.

We denote by Oσ the orbit of Pn indexed by σ ⊂ {0, 1, . . . , n}.Question: For X ⊂ T n, let X be the closure of X in Pn. Given σ ( {0, 1, . . . , n},does X intersect Oσ?.

As before, then answer depends on the configuration of trop(X) ⊂ Rn. We willreduce this calculation to one in An+1 that uses Theorem 16.8, by using the notionof the affine cone of X.

Let T n = {(t0, t1, . . . , tn) : ti ∈ k∗}. Note that we have the short exact sequence

1 → k∗ → T nπ→ T n → 1,

where 1 is the trivial group (written multiplicatively). The map k∗ → T n is given byt 7→ (t, t, . . . , t), and the map π : (t0, . . . , tn) 7→ (t1/t0, . . . , tn/t0). This short exactsequence tropicalizes to

0 → span(1) → Rn+1 trop(π)→ Rn → 0,

where 1 is the vector (1, 1, . . . , 1) ∈ Rn+1, the first map is the inclusion, and trop(π) :(w0, . . . , wn) → (w1 − w0, . . . , wn − w0).

GivenX ⊂ T n, the affine cone overX is X = π−1(X). The map π∗ : k[x±11 , . . . , x±1

n ] →k[x±1

0 , . . . , x±1n ] is given by π∗(xi) = xi/x0 for 1 ≤ i ≤ n. If X = V (I) ⊂ T n, then

X = V (π∗(I)) ⊂ T n. Note that 1 lies in the lineality space of trop(X), and thattrop(X) = trop(X)/1.Example: Let X = V (x1 + x2 + 1) ⊂ T 2. Then X = V (x1/x0 + x2/x0 + 1) =V (x1 + x2 + x0) ⊂ T 2.

Let X be the closure of X in An+1. Recall that Pn = (An+1 \ 0)/k∗. Then

X = (X) \ 0)/k∗. Thus X intersects the T n-orbit indexed by σ ( {0, . . . , n} ifand only if the preimage trop(X) of trop(X) intersects the relative interior of theappropriate pos(ei : i ∈ σ). We can also consider these sets in Rn. Note that theimage e0 of e0 in Rn is −

∑ni=1 ei. Then X intersects the T n-orbit indexed by σ if

and only if trop(X) ∩ relint(pos(ei : i ∈ σ)) 6= ∅.Example: Figure 43 shows the fan in R2 whose cones are the sets pos(ei : i ∈ σ) forσ ( {0, 1, 2}. Thus X∩Oσ 6= ∅ if and only if trop(X) intersects the relative interior ofthe appropriate cone. For an example, consider the variety X = V (x1 +x2 +1) ⊂ T 2

analyzed above.Note that T n also acts on An by (t1, . . . , tn) · (x1, . . . , xn) = (t1x1, . . . , tnxn), and

Theorem 16.8 gave conditions for the closure X in An of a variety X ⊂ T n tointersect each orbit. The sets An and Pn are examples of toric varieties. These are

62 DIANE MACLAGAN

image of pos((0, 0, 1))

image of pos((0, 1, 0))

image of pos((1, 0, 0)).

Figure 43.

varieties that contain a dense copy of T n, and have an action of T n on them extendingthe action of T n on itself. They are (up to the technical notion of normalization)described by a polyhedral fan Σ ⊂ Rn, the cones of which index T n-orbits. If X isthe closure of X ⊂ T n in a toric variety with fan Σ, then X intersects the T n-orbitindexed by the cone σ ∈ Σ if and only if trop(X) intersects the relative interior of σ.

The following proposition was also mentioned in class, and is used in the proof ofthe general case of the Fundamental Theorem.

Proposition 16.9. Let X ⊂ T n be an irreducible variety of dimension d. Then formost choices of projection φ : T n → T d+1. the image φ(X) is a hypersurface in T d+1.

Here by “most” we mean for a Zariski-open set of choices for the matrix U de-scribing φ.

Proof. We first note that since X is irreducible, φ(X) is irreducible of dimension at

most d for any choice of projection φ. To see irreducibility, note that if φ(X) = Y1∪Y2

for Y1, Y2 ( φ(X), then X = X1 ∪X2 with Xi = φ−1(Yi) ∩X for i = 1, 2. Then bythe irreducibility of X without loss of generality we have X = X1, so X ⊆ φ−1(Y1),

and thus φ(X) ⊆ Y1, contradicting Y1 ( φ(X).Since X is irreducible of dimension d and the generators of I have coefficients in k,

trop(X) is a pure d-dimensional fan in Rn. Choose a d-dimensional cone σ ∈ trop(X),and choose an n × (d + 1) rank d + 1 matrix U with ker(U) ∩ span(σ) = 0. Then

{UTw : w ∈ σ} is a d-dimensional cone in Rd+1, so trop(φ(X)) has dimension at

least d, and thus φ(X) has dimension at least d. Since X has dimension d, φ(X) hasdimension at most d, and thus is a hypersurface in T d+1. �

17. Exercises

(1) Let I be an ideal in k[x±11 , . . . , x±1

n ] generated by linear forms, and let C bethe set of circuits of I. Recall that a circuit is a polynomial f =

∑ni=1 aixi ∈ I

with support {i : ai 6= 0} minimal with respect to inclusion. Show that C isa tropical basis for I.

(2) Let τ be a trivalent tree with n leaves. Show that τ has 2n− 3 edges.


(3) Let φ : T n → Tm be a morphism of tori, given by φ(t)i = tui , and let U bethe n×m matrix with columns u1, . . . ,um. Show that φ is surjective if andonly if U has rank m.

(4) Let X = V (x + y + z + 1) ⊆ T 3. For each of the following maps of tori

φ : T 3 → Tm compute φ(X) ⊂ Tm , and verify that trop(φ(X)) = {UTw :w ∈ trop(X)}, where U is the 3 × m matrix with columns u1, . . . ,um forφ(t)i = tui .

(a) φ : T 3 → T 2 given by U =

1 00 10 0

.

(b) φ : T 3 → T 2 given by U =

1 12 −11 0

.

(c) φ : T 3 → T 4 given by U =

1 0 0 00 1 0 00 0 1 0

.

(5) For which of the following X ⊂ T 3 is 0 ∈ X ⊂ A3? Compute trop(X) andverify the statement of the Theorem.(a) X = V (3x2 + 3xy + 2x+ y);(b) X = V (3x2 + 3xy + 2x+ y + 1);(c) X = V (x+ y + z, x+ 2y);(d) X = V (x+ y + z + 1, x+ 2y + 3z);

(6) (a) Check that T 2 ⊂ P2 is a Zariski-dense subset, and that there is an actionof T 2 on P2 that extends the action of T 2 on itself.

(b) List the orbits of T 2 in P2.(c) Let X = V (x + y + x2y + xy2) ⊂ T 2, and let X be the closure of X in

P2. How does X intersect each of the T 2-orbits of P2?(d) Draw trop(X) ⊂ R2. How does this relate to your previous answer?

18. Lecture 16

Summary of Class:We begin this lecture by summarizing what we have done in this class.We have constructed a tropical variety trop(X) associated to a variety X ⊂ T n

with X = V (I) for I ∈ K[x±11 , . . . , x±1

n ]. This was defined by

trop(X) =⋂f∈I

trop(V (f)),

where for f =∑

v∈Zn cvxv the set trop(V (f)) is the set of w ∈ Rn for which the

tropicalization trop(f)(w) = min(val(cv) +w · v) is achieved at least twice. We havethe Fundamental Theorem of Tropical Algebraic Geometry:

Theorem 18.1. Let X = V (I) ⊂ T n for I ⊂ K[x±11 , . . . , x±1

n ]. Then the followingsets coincide:

(1) trop(X);(2) The closure in Rn of {w ∈ (im val)n : inw(I) 6= 〈1〉}.

64 DIANE MACLAGAN

(3) The closure in Rn of {val(y) : y ∈ X}.

The last part of the Fundamental Theorem means we can think of a tropical varietyas a combinatorial shadow of the variety X ⊂ T n. We saw that tropical varieties havethe following structure:

Theorem 18.2 (Structure Theorem for Tropical Varieties). Let X ⊂ T n be anirreducible variety of dimension d. Then trop(X) is the support of a balanced weightedpolyhedral complex that is pure of dimension d that is connected in codimension one.

We saw this in examples we computed with gfan, and in the specific explicitexamples of linear varieties and G(2, n).

In the spirit of the “combinatorial shadow” philosophy, the (somewhat philosoph-ical) question guiding work in this form of tropical geometry is:Question: Which invariants of X can be computed from trop(X)?

For example, we see from the main structure theorem that dim(X) = dim(trop(X)),so if we know the tropical variety we know the dimension. An answer to the guidingphilosophical question is a current area of active research.

Recall that a variety P is a toric variety if P contains a dense copy of T n, andthere is an action of T n on P extending the action of Tn on itself. A polyhedral fanΣ defines a toric variety Pσ. The cones of Σ index the T n-orbits of Pσ. Examplesinclude T n, An, Pn, and P1 × P1.

Often we are interested not in some X ⊂ T n but in a projective variety Y ⊂ Pmfor some m. Tropical geometry can be useful if this occurs in the following way.

(1) Choose X ⊆ Y and an embedding X ⊂ T n for some n.(2) Let trop(X) be the fan Σ ⊂ Rn, and let PΣ be the toric variety with fan Σ.(3) Let X be the closure of X in PΣ.(4) If Y ∼= X then we can hope to learn about Y = X from trop(X).

A variety Y that arises in this fashion is called a tropical compactification of X.This was first considered in the work of Tevelev [Tev07] with extra conditions on thefan structure on Σ, which guarantees nicer properties for the geometry of X.

One of the main invariants of Y = X we can hope to learn from trop(X) is theChow ring of Y , which is the algebraic version of the cohomology ring. This areaof algebraic geometry is known as intersection theory. Those unfamiliar with thecohomology ring should spend some time with an algebraic topology book, such as[Hat02] (freely electronically available from the author’s webpage). The enumerationof curves problem we will discuss next is such a form of intersection problem.

Other aspects of tropical geometryThis philosophy of obtaining information about X ⊂ T n, or a projective com-

pactification X, from the tropical variety trop(X) is only one portion of currentresearch in tropical geometry. Some samples of this philosophy are found in thework of Speyer [DS05], Payne [Pay07], Tevelev [Tev07], and Hacking, Keel, andTevelev [HKT06]. The key ideas behind the Fundamental Theorem arose from thework of Kapranov, which appears in [EKL06]. We emphasize that these referencesare just a (nonrepresentative) sample of the literature.


trop(X1)

=⋃

Figure 44.

We now give some ideas of other aspects of tropical geometry. This is only a briefoverview, and references will just be a nonrepresentative sample.

Abstract tropical varietiesWe can define a tropical variety to be a balanced weighted polyhedral complex

Σ, and then define tropical versions of algebraic geometry invariants by analogy withtheir classical definitions. An example of this includes defining the dimension to bethe dimension of the complex Σ. Current work (see [LAJR07]) is developing a versionof intersection theory in this context. Another possibility would be a tropical versionof irreducible components. This creates difficulties, though, as the following exampleshows.Example: Let X1 = V ((x+ y+ 1)(x+ y+ xy)) = V (x+ y+ 1)∪ V (x+ y+ xy) ⊂T 2. Then trop(X1) = trop(x + y + 1) ∪ trop(x + y + xy) is shown in Figure 44.Let X2 = V ((x − 1)(y − 1)(x − y)) = V (x − 1) ∪ V (y − 1) ∪ V (x − y). Thentrop(X2) is the union of the lines w1 = 0, w2 = 0 and w1 = w2, so equals trop(X1).Since we can’t write any of these last three lines as the union of smaller balancedweighted polyhedral complexes, this means that trop(X1) = trop(X2) cannot bewritten uniquely as the union of “irreducible” tropical varieties, so a naive definitionof irreducible components of these “abstract” tropical varieties does not work.

Similarly, we will see later in the week that there are three different notions of theranks of a matrix over the tropical semiring. One of the motivations for understandingwhich geometric invariants make sense for these abstract tropical varieties is that itindicates which properties we can expect combinatorial proofs and interpretationsof. This can be thought of as the synthetic approach to tropical geometry. Someexamples can be seen in the work of Gathmann and his students, such as [GK08],[LAJR07].Warning: Note every balanced weighted polyhedral complex in Rn is trop(X) forsome X ⊂ T nK . See the example of Mikhalkin in [DS05, Figure 5.1].Open Question: Characterize those balanced weighted polyhedral complexes inRn that are trop(X) for some X ⊂ T nK .

More general tropical geometryAnother vital branch of tropical research can be broadly classified as attempting

to do all geometry, not merely algebraic geometry, over the tropical semifield. For

66 DIANE MACLAGAN

example, there should be tropical manifolds, and complex and differential geometry inthis setting. This the approach currently taken by Mikhalkin to establish the basics oftropical geometry. See [Mik06], [GM07], [GM] for some expositions of this approach.This also leads to intersections and applications for real algebraic geometry; see, forexample, [IKS05]. See [IMS07] for more in this line.

Other aspectsAnother major school of tropical geometry revolves around the approach of Gross

and Siebert to mirror symmetry using tropical geometry; see [GS06] and its refer-ences.

Part of the excitement of this field is that new areas and applications are constantlyarising. For example, Gubler [Gub07] recently used the connections to rigid analyticgeometry and Berkovich spaces for applications in number theory. Finally, as wellas the emerging tropical geometry community, there is also research in max-plusalgebras in control theory and related topics, which is a more established field.

These references are only a sample of the exciting research happening in tropicalgeometry. For more, google “tropical geometry”, or put “tropical” in the “anywhere”search box at front.math.ucdavis.edu or mathscinet.

19. Lectures 17 and 18

In these last two lectures we describe how to use tropical techniques to countrational curves in P2 passing through a fixed number of points.

Definition 19.1. A curve C = V (f) ⊂ P2 has degree d if f ∈ C[x0, x1, x2] ishomogeneous of degree d. An irreducible curve C is rational if there is a map froman open set U ⊂ P1 to C given by

φ([t0 : t1]) = (p0(t0, t1) : p1(t0, t1) : p2(t0, t2))

where pi is a homogeneous polynomial of degree d for 0 ≤ i ≤ 2. The open setU ⊂ P1 is the set of [t0 : t1] for which at least one of p0(t0, t1), p1(t0, t1), and p2(t0, t1)are nonzero.

A curve of degree one is a line in P2, which is rational.Exercise: Check that every curve in P2 of degree two is rational.Question: Given n general points p1, . . . , pn ∈ P2, how many rational curves ofdegree d pass through all n points?

This question clearly needs to be clarified before a clear answer can be given. Forsmall n (such as n = 1) there will be an infinite number of curves. For example,there are infinitely many lines through any given point in P2. For large n if the piare not chosen specially, then there are no curves. For example, if p1, p2, p3 are threepoints in P2 that do not lie on a line, then there are no curves of degree one (lines!)passing through all three points. However given any two distinct points in P2, thereis a unique line passing through them, so for d = 1 and n = 2, with the notion of“general” being “distinct”, the question has answer one.

When d = 2, we claim that there is also a unique curve of degree two passingthrough five general points in P2. To see this, let

F = ax20 + bx0x1 + cx0x1 + dx2

1 + ex1x2 + fx22.


The variety C = V (F ) is a curve as long as one of a, . . . , f is nonzero, and F andλF define the same curve, so a choice of conic corresponds to a point [a : b : c : d :e : f ] ∈ P5. We observed in the exercise above that all curves of the form V (F ) arerational, so we need to show that given five sufficiently general points in P2 there is aunique point in P5 for which the corresponding V (F ) passes through all five points.Since we are requiring that the five points be general, we may assume that the firstcoordinate is nonzero, so they have the form P5 = {[1 : ui : vi] : 1 ≤ i ≤ 5}. So weneed the equation

1 u1 v1 v21 u1v1 v2

1

1 u2 v2 v22 u2v2 v2

2

1 u3 v3 v23 u3v3 v2

3

1 u4 v4 v24 u4v4 v2

4

1 u5 v5 v25 u5v5 v2

5

abcdef

= 0

to have a one-dimensional solution space for most choices of P5. This means thatthe 5× 6 coefficient matrix must have rank five for most choices of five points. Theideal of 5× 5 minors of the coefficient matrix is principal, and generated by a singlepolynomial of degree six in C[u1, . . . , u5, v1, . . . , v6], so for any choice of P5 with thispolynomial nonzero there is a unique rational curve of degree two passing throughthe points in P5.

Note that this means that for a general set of four points in P2 there are are aninfinite number of degree two rational curves passing through the points, and for ageneral set of six points in P2 there are no degree two rational curves passing throughall of the points.Claim: For general d, there are a finite number of rational curves of degree dpassing through 3d− 1 general points in P2.Idea of proof: A rational curve of degree d in P2 is determined by three polynomialsp0, p1, p2, which have the form

∑dj=0 aijt

j0td−j1 for 0 ≤ i ≤ 2, for a total of 3d + 3

parameters. Since the image is in P2, this over-counts by one. Since we only careabout the image of the curve, not the parameterization, this also over-counts by thedimension of AutP1, which is three. This means that the set of rational curves isdetermined by 3d + 3 − 1 − 3 = 3d − 1 parameters. Thus forcing the curve to passthrough 3d− 1 general points in P2 will guarantee a finite number of solutions.

Definition 19.2. LetNd be the number of irreducible rational curves passing through3d− 1 general points in P2.

Example: We saw above that N1 = N2 = 1. The numbers N3 and N4 werecomputed in the nineteenth century.

Theorem 19.3 (Kontsevich). For d > 1 the numbers Nd obey the following recursion:

Nd =∑

dA+dB=d,dA,dB>0

(d2Ad

2B

(3d− 4

3dA − 2

)− d3

AdB

(3d− 4

3dA − 1

))NdA

NdB.

Note that this describes Nd in terms of smaller d, so knowing N1 = 1 determinesall larger Nd.

68 DIANE MACLAGAN

Example: To compute N2, the only decomposition is dA = dB = 1. Then

N2 = (1212

(2

1

)− 131

(2

2

))(1)(1) = 2− 1 = 1.

To compute N3 we need to consider the pairs (dA, dB) ∈ {(1, 2), (2, 1)}. Thus

N3 =(1222

(5

1

)− 13(2)

(5

2

))(1)(1)

+ (2212

(5

4

)− 23(1)

(5

4

))(1)(1)

=20− 20 + 20− 8

=12

Tropical VersionWe now outline how to prove Theorem 19.3 using tropical methods. Our sketch

follows closely the version given in [HM06], which is based on [GM08] and [Mik05].The idea is to define a tropical analogue N trop

d of Nd, and show that this equals Nd.We then show that N trop

d obeys the Kontsevich recursion Theorem 19.3.

Definition 19.4. A tropical rational curve of degree d is a one-dimensional balancedweighted polyhedral complex for which

(1) the unbounded rays point in the directions (1, 0), (0, 1), and (−1,−1);(2) there are d unbounded rays pointing in each of these directions (counted with

multiplicity);(3) and the underlying graph of the polyhedral complex has no cycles.

Example:Figure 45 shows contains some examples of tropical rational curves of degrees two,

two, and three. Figure 46 contains some one-dimensional tropical varieties that arenot tropical rational curves of degree d for some d. For the first variety, the problemis that the underlying graph contains a cycle. For the second, there are unboundededges not pointing in one of the three prescribed directions.

Note that there is a unique tropical rational curve of degree one through two pointsin R2 unless they lie on the same vertical, horizontal, or slope-one line.Exercise: There is a unique tropical rational curve of degree two through mostsets of five points in R2.

One way to construct a tropical rational curve of degree d is to take a curve of theform X = V (f) ⊂ T 2

C{{t}}, where f ∈ C{{t}}[x±1, y±1] has the form∑cijx

iyj wherecij 6= 0 if and only if i, j ≥ 0 and i+j ≤ 2. Then, as computed in the first exercise set,trop(X) is a weighted balanced polyhedral complex with d unbounded rays pointingin the prescribed directions. In [DES07] Speyer shows that every tropical rationalcurve of degree d in R2 arises in this fashion.

Definition 19.5. A rational tropical curve of degree d is trivalent if the underlyinggraph is trivalent. Let C be a rational tropical curve of degree d, and let V be a


2

2

2

Figure 45.

Figure 46.

vertex of C. Let v1,v2,v3 be the smallest integral vectors pointing along the threerays leaving V , and let µ1, µ2, µ3 be the multiplicities of the corresponding polyhedra.The multiplicity of V is the absolute value of the determinant of the 2×2 matrix withcolumns two of the µivi. The balancing condition guarantees that this is independentof the choice. The multiplicity of C is the product of the multiplicity of all verticesin C.

Example : The multiplicity of the first two tropical rational curves shown in Fig-ure 47 is one. The multiplicity of the third curve is 8 = (2)(4), since the multiplicityof the bottom vertex is 4, and the multiplicity of the vertex above is 2, while theother two vertices have multiplicity one.

Definition 19.6. Fix P = {p1, . . . , p3d−1} ⊂ R2. Then N tropd (P) is the number of

tropical rational curves counted with multiplicity passing through p1, . . . , p3d−1.

Proposition 19.7. There is a Zariski-open set U ⊂ (R2)3d−1 for which N tropd (P) is

constant for P = {p1, . . . , p3d−1} with (p1, . . . , p3d−1) ∈ U .

70 DIANE MACLAGAN

2

22

2

4

Figure 47.

Figure 48.

Definition 19.8. A set P = {p1, . . . , p3d−1} with (p1, . . . , p3d−1) ∈ U for the set Uof Proposition 19.7 is said to be in tropical general position. Let N trop

d = N tropd (P)

for any P in tropical general position

Theorem 19.9 (Mikhalkin Correspondence Theorem). We have equality

N tropd = Nd.

The idea of the proof of Theorem 19.9 is as follows.The amoeba of a variety X ⊂ (C∗)2 is the set {(log(|x1|, log(|x2|)) : (x1, x2) ∈

X} ⊂ R2.Example: Let C = V (x1 + x2 + 1) ⊂ T 2

C = {(a,−1 − a) : a ∈ C∗, a 6= −1}. Thenthe amoeba of C is shown in Figure 48.

When C ⊂ T 2C has higher degree d, its amoeba generically has d “tentacles” point-

ing in each direction.If we replace log by logt in the definition of the amoeba, and let t → ∞, then

the amoeba gets thinner and thinner, and in the limit approaches a tropical curve ofdegree d. Given a set of 3d − 1 points we can count the number of rational curvesof degree d passing through these points, or the number of tropical rational curvesof degree d passing through (roughly) the logs of the points. The multiplicity of atropical rational curve counts how many of these rational curves in T 2 limit to thetropical curve.


2

c

c

a b

2

1

φ

a

c

1 = 2

b

1

a b

2

c

φ

a

1

b

Figure 49.

By Theorem 19.9, in order to prove Theorem 19.3, it suffices to show that thenumbers N trop

d satisfy the same recursion. We show this by using the standardenumerative combinatorics trick of arguing that the equation(5)

Nd+∑

dA+dB=d,dA,dB>0

d3AdB

(3d− 4

3dA − 1

)NdA

NdB=

∑dA+dB=d,dA,dB>0

d2Ad

2B

(3d− 4

3dA − 2

)NdA

NdB

counts the same objects in two different ways. These objects are parameterizedtropical rational curves with n = 3d marked points.

Definition 19.10. A parameterized tropical rational curve of degree d with nmarkedpoints is a map φ : Γ → R2 where Γ is a tree with 3d+ n leaves such that

(1) n of the leaves of Γ are labelled 1, . . . , n,(2) each non-leaf edge e of Γ comes with a weight de,(3) φ(Γ) is a rational tropical curve of degree d,(4) the image of the labelled leaves of Γ are contracted to points, and(5) the images of the nonlabelled leaves of Γ extended to unbounded rays.

Example: Two examples of parameterized rational curves of degree one with twomarked points are shown in Figure 49. We have labelled the “unlabelled” edges a, b,and c so their image under φ is clear, but this is not part of the data of φ. Note thata non-leaf edge of Γ is contracted in the second example.

Definition 19.11. Given a parameterized tropical curve of degree d with n ≥ 4marked points φ : Γ → R2, we define the forgetful map f as follows. Let Γ′ be thesmallest connected subtree of Γ containing the leaves 1, 2, 3 and 4. Then Γ′ consistsof two pairs of leaves which are connected by a path of non-leaf edges. The image ofthe forgetful map f(φ) is then the phylogenetic tree Γ with four leaves obtained byturning this path of edges into one edge with length the combined weights.

Example: An example of the forgetful map applied to a parameterized tropicalcurve of degree one is shown in Figure 50.

72 DIANE MACLAGAN

f

1

4

2

3

5

1

32

5

φ

1

4

2

3

a

b

a + b

Figure 50.

We will count the number Md(p1, . . . , p3d−1; Γ) of parameterized tropical rationalcurves φ : Γ → R2 of degree d with n = 3d marked points such that

(1) The labelled edges 3, . . . , n get mapped to p1, . . . , p3d−1;(2) The labelled edge 1 gets mapped to a point on the line x = (p1)1;(3) The labelled edge 2 gets mapped to a point on the line y = (p1)2;(4) The image f(φ) of the forgetful map is equal to Γ.

Proposition 19.12. The number Md(p1, . . . , p3d−1; Γ) is constant for a general choiceof p1, . . . , p3d−1 ∈ R2 (tropical general position) and for any choice of four-vertexphylogenetic tree Γ. We thus denote it by Md.

The idea of the proof is then to choose p1, . . . , p3d−1 in tropical general position,and then choose two different trees Γ at which to evaluate Md to obtain equation 5.

The proof will use the tropical version of Bezout’s theorem. Recall that Bezout’stheorem in the plane says that if the variety V (f1, f2) ⊂ P2 is finite, when f1, f2

are homogeneous polynomials in C[x0, x1, x2] of degree d1 and d2 respectively, thenV (f1, f2) consists of d1d2 points, counted with multiplicity. The (weak) tropicalversion of this states that if C1, C2 are two tropical curves in R2, of degrees d1 and d2

respectively, that intersect in a finite number of points, then that intersection consistsof d1d2 points counted with multiplicity. An example of two tropical rational curvesof degree two intersecting in four points is shown in Figure 51.

We can now outline the proof of Equation 5. First choose a phylogenetic tree Γwith four labelled leaves that looks like the one on the left of Figure 52, with thelength a of the bounded edge large. Let φ : Γ → R2 be a parameterized rationaltropical curve of degree d with n = 3d labelled points whose image under the forgetfulmap is Γ. Then one of two situations occur. The first is that the two leaves labelled1 and 2 are adjacent to the same vertex in Γ, so the images of these two leaf edges in


Figure 51.

b

1

2

3

4

1

3

2

4

a

Figure 52.

φ

Γ

Figure 53.

R2 is the same. This means that φ(Γ) is a tropical rational curve of degree d in R2

with φ(1) = φ(2) = p1, and φ(i) = pi−1 for 3 ≤ i ≤ n. The number of such images isNd. Such maps φ are determined by their image, so there are Nd of such φ.

The second possibility is that 1 and 2 are not adjacent to the same vertex in Γ.Then it can be shown (see [HM06, Remark 7.16]) that there is a contracted edge ofΓ, which leads to the image φ(Γ) being reducible. See Figure 53 for an example withno marked points. This means it is the union of two tropical rational curves CA andCB, of degrees dA and dB respectively, where 1 and 2 live on CA, 1 lives on the linex = (p1)1, 2 lives on the line y = (p1)2, 3 and 4 live on CB, 3dA − 1 of {5, ,n} live onCA, and φ(i) = pi−1 for 3 ≤ i ≤ n. There are

(3d−4

3dA−1

)choices of the 3dA − 1 points

74 DIANE MACLAGAN

that live on CA. By Bezout the tropical curve CA intersects the line x = (p1)1 (whichis itself a tropical line) dA times, and similarly for the line y = (p1)2. There are thusdA choices for the image of 1, and dA choices for the image of 2. Finally, the curvesC1 and C2 intersect in dAdB points, so there are that many choices for the locationof the contracted edge of Γ. This gives a total of

d3AdB

(3d− 4

3dA − 1

)NdA

NdB

choices for such φ, so

(6) Md = Nd + d3AdB

(3d− 4

3dA − 1

)NdA

NdB.

Alternatively, choose a phylogenetic tree Γ with four labelled leaves that looks likethe one on the right of Figure 52, where the length b of the bounded edge is large. Inthis case we cannot have φ(1) = φ(2), as that would mean that two other φ(i) wouldhave to coincide, so two of the pi would have to coincide, which is ruled out by thetropical general position hypothesis. So we are in the second case above, where theimage φ(Γ) is a reducible tropical rational curve. In this case we have 1 lying on thecurve CA of degree dA, and 2 lying on the curve CB of degree dB. The point 3 liveson CA and the point 4 lives on CB. There are 3dA− 2 of the points {5, . . . , n} livingon CA. There are thus

(3d−4

3dA−2

)choices for these points, dA choices for the image of 1,

dB choices for the image of 2, and dAdB choices for the image of the contracted edgeof Γ. This gives

(7) Md = d2Ad

2B

(3d− 4

3dA − 2

)NdA

NdB.

Combining Equations 6 and 7 we obtain equation 5, and thus have the outline of thetropical proof of Kontsevich’s formula.

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